NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Short Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Short Question And Answers

Question 1. How much time would It take to distribute one Avogadro number of wheat grains If 10¹⁰grains are distributed each second?
Answer:

1010 grains are distributed in 1 sec. So, the time required to distribute 6.022 × 10²³grains would be

= \(\frac{1 \times 6.022 \times 10^{23}}{10^{10}}=6.022 \times 10^{13} \sec\)

= \(696.990 \times 10^6 \text { days }\)

= \(1.909563 \times 10^6 \text { years }\)

Question 2. Calculate the total pressure In a mixture of 8 g of dioxygen and 4 g of dihydrogen confined In a vessel of 1 dm³ at 27°C, R = 0.083 bar .dm³. K^-1mol^-1
Answer:

⇒ \(\left(\frac{8}{32}+\frac{4}{2}\right)=2.25 \mathrm{~mol}\) Molar mass of O₂and H₂ are 32 and 2 g mol-1 respectively]

As given: V = 1dm3 and T = (273 + 27)K = 300K

∴ \(P=\frac{n R T}{V}=\frac{2.25 \times 0.083 \times 300}{1}=56.025 \text { bar. }\)

Question 3. What will the nature of the PV vs P graph be for a real gas at Boyle temperature?
Answer:

AOS, At Boyle temperature the value of PV, particularly in the low-pressure region, becomes constant. Thus, the gas shows ideal behaviour. So, at this temperature, the plot pv of PV against P will give a straight line parallel to the P-axis

Question 4. Of the following types of velocity, which one has the highest value and which one has the lowest value at a given temperature?
Answer:

At a given temperature; the average velocity, the root mean square velocity, and the most probable velocity of the molecules of gas with molar mass M are given by average velocity

⇒  \((\bar{c})=\sqrt{\frac{8 R T}{\pi M}}\) and most Probable velocity

⇒ \(\left(c_m\right)=\sqrt{\frac{2 R T}{M}}\)

These expressions indicate that at a given temperature a given gas has the highest value and cm has the lowest value.

Question 5. Why are the deviations from the ideal behaviour of CO₂ and CH₄ greater than those of H₂ and He?
Answer:

The Molar masses of CO₂ and CH₄ are greater than those of H₂ and He. Hence, the intermolecular attractive forces in CO₂ and CH₄ are also greater in magnitude than those in H₂ and He. This results in a greater deviation from ideal behaviour for CO₂ and CH₄ than that for H₂ and He.

Question 6. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen
Answer:

Let the weight of the mixture be w g. So, in the mixture, the weight of H₂ gas = 0.2 w g and that of O₂ gas =(w-0.2w) =0.8 w g

⇒ \(0.2 w g \text { of } \mathrm{H}_2=\frac{0.2 w}{2}=0.1 \times w \mathrm{~mol} \text { of } \mathrm{H}_2\)

⇒ \(0.8 \mathrm{wg} \text { of } \mathrm{O}_2=\frac{0.8 w}{32}=0.025 \times w \mathrm{~mol} \text { of } \mathrm{O}_2\)

So, the partial pressure of H₂ in the mixture = JCH x total pressure of the mixture =0.8 × 1 bar = 0.8 bar

Question 7. What would be the SI unit for the quantity \(\frac{P V^2 T^2}{n}\)
Answer:

In the SI system, the units of P, V, T and n are N-m~2, m3, K and mol respectively So, the SI unit for the quantity

⇒  \(\frac{V^2 T^2}{n}\) will be \(\frac{\mathrm{N} \cdot \mathrm{m}^{-2} \times\left(\mathrm{m}^3\right)^2 \cdot \mathrm{K}^2}{m o l} \text { i.e., } \mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{K}^2 \cdot \mathrm{mol}^{-1}\)

Question 8. The critical temperature for carbon dioxide and methane arc is 31.1 °C and -81.0°C respectively. Which of these has stronger Interiuolecuhir forces and why?
Answer:

The critical temperature of gas Is a measure of the intermolecular forces of attraction In the gas. A gas with stronger intermolecular forces of attraction has a higher critical temperature. Therefore, CO₂., gas possesses stronger intermolecular forces of attraction than CH₄ gas because CO₂ has a higher critical temperature than CH₄.

Question 9. A 2L flask contains 0.4 g O₂ and 0.6 g H₂ at 100°C. Calculate the total pressure of the gas mixture in the flask. 4
Answer:

Total no. of moles of O₂ & H₂ in flask \(=\frac{0.4}{32}+\frac{0.6}{2}=0.3125\)

∴ \(P=\frac{n R T}{V}=\frac{0.3125 \times 0.082 \times 373}{2}=4.78 \mathrm{~atm}\)

Question 10. Arrange CO₂, SO₂ and NO₂ gases in increasing order of their rates of diffusion under the same condition of temperature and pressure with reason.
Answer:

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Hence increasing order of rates of diffusion is:

⇒ \(r_{\mathrm{SO}_2}<r_{\mathrm{NO}_2}<r_{\mathrm{CO}_2} .\)

Question 11. Explain the nature of the graphs of log P versus log V and logy versus log T. What are the units of the van der Waals constants ‘a’ and ‘b’?
Answer:

The plot of log P versus log V indicates a straight line with a negative slope (-1). The plot of log V versus log indicates a straight line with positive slope (+1)

Unit of is atm. L² mol¯² a unit of ‘b’ is L-mol^-1

Question 12. Prove that c = [E= total kinetic energy molecules of 1 mol of a gas, M=molar mass of the gas, arms = root mean square velocity of gas molecule]
Answer:

We known \(c_{r m s}=\sqrt{\frac{3 R T}{M}} \quad \text { or, } c_{r m s}=\sqrt{\frac{2}{M} \times \frac{3}{2} R T}\)

The total kinetic energy of the molecules of lmol gas, \(E=\frac{3}{2} R T\)

∴ \(c_{r m s}=\sqrt{\frac{2 E}{M}} \quad \text { (Proved) }\)

Question 13. It is easier to liquefy a gas with a higher critical temperature—explain.
Answer:

The critical temperature of a gas reflects the strength of intermolecular attractive forces in the gas. A gas with a higher value of critical temperature possesses a stronger intermolecular force of attraction. Now, the stronger the intermolecular forces of attraction in a gas, the easier it is to liquefy the gas. Therefore, a gas with a higher critical temperature can be liquified easily.

Question 14. If P is plotted against 1/ V for 1 mol of an ideal gas at 0°C, then a straight line passing through the origin is obtained. What is the slope ofthe straight line?
Answer:

For 1 mol of an ideal gas, PV = RT or, P = RT/V At 0°C or 273K, P = 273R/V.

Hence, the slope of the P vsl/V plot is =273 R

Question 15. The number of molecules in an ideal gas with a volume of V at pressure P and temperature T is ‘n Write down the equation of state for this gas
Answer:

No. of mole of the gas

⇒ \(\frac{\text { Total no. of molecules of the gas }}{\text { Avogadro’s no. }}=\frac{n}{N}\)

∴ The equation of state \(P V=\frac{n}{N} R T\)

Question 16. On what factors does the value of the total kinetic energy of the molecules in a gas depend?
Answer:

The total kinetic energy of the molecules of 1 mol gas \(=\frac{3}{2} R T\)

The total kinetic energy of n mole gas \(=n \times \frac{3}{2} R T\)

Therefore, the total kinetic energy of any gas depends upon the temperature and the quantity ofthe gas.

Question 17. At a definite temperature, the total pressure of a gas mixture consisting of three gases A, B and C is P. If the number of moles of A, B and C are 2, 4 and 6, respectively, then arrange these gases in increasing order of their partial pressures
Answer:

The mole fractions of A, B and C in the gas mixture

⇒ \(x_{\mathrm{A}}=\frac{2}{12}=\frac{1}{6}, x_B=\frac{4}{12}=\frac{1}{3} \text { and } x_C=\frac{6}{12}=\frac{1}{2}\)

Therefore The Partial Pressure Of Gas A ⇒  \(A, p_A=\frac{P}{6}, p_B=\frac{P}{3}\)

⇒  \(p_C=\frac{P}{2}\).

Hence PA<PB<PC.

Question 18. Under identical conditions of temperature and pressure, it takes time t1 for the effusion of VmL of N₂ gas through a porous wall and time t₂ for the effusion of the same volume of O₂ gas through the same porous wall. Which one is greater, t₁ or t₂?
Answer:

At constant temperature and pressure, the rates of effusion of different gases are inversely proportional to the square roots of their molar masses. As the molar mass of N₂ is smaller than that of CO₂, at the given conditions the time required for the effusion of VmL of N₂ gas will be less than that required for V mL CO₂ of gas. Hence, t₁<t₂.

Question 19. Arrange the following gases in the increasing order of their densities at STP: H₂, air, CO₂.
Answer:

The density of a gas at constant temperature and pressure \(d=\frac{P M}{R T}.\)

Thus, at a particular temperature and pressure, the density of the gas is directly proportional to the molar mass of the gas. The order of the molar mass of the given gases is: M_H2< M_air < MCO₂.

Therefore, the order of densities of these gases at STP would be, d_H2 < d_air < dCO₂.

Question 20. At a constant temperature, the pressures of four gases A, B, C, and D are 0.2 atm, 250torr, 26.23k Pa, and 14.2 bar, respectively. Arrange them according to their increasing pressure.
Answer:

⇒  \(A: 0.2 \mathrm{~atm}, B: 250 \mathrm{torr}=\frac{250}{760} \mathrm{~atm}=0.33 \mathrm{~atm}\)

⇒ \(A: 0.2 \mathrm{~atm}, B: 250 \mathrm{torr}=\frac{250}{760} \mathrm{~atm}=0.33 \mathrm{~atm}\)

⇒ \(D: 14.2 \mathrm{bar}=\frac{14.2}{1.013}=14.02 \mathrm{~atm}\)

∴ A<C<B<D

Question 21. Among the four quantities—mass, pressure, temperature, and volume, which are taken to be constant in the following gas laws?

• Boyle’slaw
• Charles’law
• Gay-Lussac’slaw
• Avogadro’s law.

Answer:

• Boyle’s law: Mass and temperature ofthe gas
• Charles’ law: Mass and pressure of the gas
• Gay-Lussac’s law: Mass and volume of the gas
• Avogadro’s law: Pressure and temperature of the gas

Question 22. Why does the volume of a given mass of gas increase by decreasing its pressure at constant temperature?
Answer:

According to Boyle’s law, PV= constant for a fixed mass of gas at a constant temperature. That is, the product of P and V for a fixed mass of gas at a constant temperature is always constant. Suppose, P and V are the pressure and volume of a given mass of gas at a constant temperature.

Keeping the temperature constant, if the pressure is made to \(\frac{P}{x}\)(where x > 1 ), then the volume ofthe gas will be V × x because PV = constant

Question 23. Why does the volume of a gas increase by increasing its number of moles at a given temperature and pressure?
Answer:

According to Avogadro’s law, at a given temperature and pressure the volume (V) of a gas is directly proportional to its number of moles (n), i.e., V ∝ n. Therefore, if the number of moles of a gas is increased at a constant temperature and pressure, its volume will increase

Question 24. Will the nature of the following graphical presentations for a given mass of gas be the same?
Answer: 

P vs V atm constant temperature

For a fixed mass of a gas at a given temperature PV = constant. This relation expresses an equation of a rectangular hyperbola. Therefore, the P vs V plot for a fixed mass of gas at a given temperature will produce a curve of a rectangular hyperbola.

V vs T at constant pressure

For a fixed amount of a gas at a constant pressure, V = KxT (K = constant). This relation expresses an equation of a straight line passing through the origin. Therefore, the V vs T plot for a fixed mass of gas at constant pressure will give a straight line passing through the origin.

Question 25. Using the equation of state PV = nBT, show that at a given temperature, the density of a gas is proportional to the gas pressure P.
Answer:

P V = n R T  or, \( P V=\frac{g}{M} R T\)

M= Molar mass of gas in g.mol^-1

g= mass of gas in g]

Or \(P M=\frac{g}{V} R T[latex]

= d R T

d= g/v= density of the gas

Since, B is constant dec P, when T is constant

Question 26. At a low pressure, the van der Waals equation reduces What is the value of [latex]\left(P+\frac{a}{V^2}\right) V=R T\) compressibility factor (Z) for this case at this condition?
Answer:

⇒ \(\left(P+\frac{a}{V^2}\right) V=R T\)

or, \(P V+\frac{a}{V}=R T \quad \text { or, } \frac{P V}{R T}+\frac{a}{R T V}=1 \quad \text { or, } Z=1-\frac{a}{R T V}\)

Question 27. Rank the gases N₂, CO₂, and CH₅ in order of their increasing densities given temperature and pressure.
Answer:

We know \(d=\frac{P M}{R T}\) Thus, docM at a certain temperature and pressure.

Since, \(M_{\mathrm{CH}_4}<M_{\mathrm{N}_2}<M_{\mathrm{CO}_2}\) at a fixed temperature and pressure, \(d_{\mathrm{CH}_4}<d_{\mathrm{N}_2}<d_{\mathrm{CO}_2}\)

Question 28. Under different conditions, the following graph is obtained for an ideal gas. Mentioning A and B,- identify the conditions.
Answer:

For a fixed amount of gas at a given temperature, PV is constant. This means the value of PV will always be constant for a fixed amount of a gas at a given temperature no matter what the pressure of the gas is. Hence PV vs P or PV vs V plot will give a straight line parallel to the P-axis or V-axis, respectively. Therefore, A will be equal to PV and B will be equal to P or V.

For a fixed amount of a gas at a given pressure V – K (constant) x T, i.e., V/T = K. This relation tells us that the value of V/ T will always be constant for a fixed amount of gas at a given pressure irrespective of the value of temperature of the gas. Hence, the V/T vs T plot will give a straight line parallel to the T axis. Hence, A will be equal to V/ T and B will be equal to T.

Question 29. If a substance were to be in a gaseous state at absolute zero temperature, what would be the theoretical value of its pressure?
Answer:

⇒ \(V_t=V_0\left(1+\frac{t}{273}\right)\)

Pressure and mass of gas being constant] At absolute zero temperature, t = -273°C.

⇒ \(\text { Hence, } V_{-273^{\circ} \mathrm{C}}=V_0\left(1-\frac{273}{273}\right)=0 \text {. }\)

Since the volume of the gas is zero (0), the theoretical value of pressure will be zero (0).

Question 30. Under similar conditions of temperature and pressure, the times it takes for the effusion of the same volume of H₂, N₂, and O₂ gases through the same porous wall are t₁ t₂ and t₃, respectively. Arrange t₁, t₂, and t₃ in order of their increasing values.
Answer:

Let the volume of effused gas be the rates of effusion ofthe three gases will be as follows—

⇒ \(\text { For } \mathrm{H}_2: \frac{V}{t_1} \propto \frac{1}{\sqrt{M_{\mathrm{H}_2}}} \cdots[1] ; \text { For } \mathrm{N}_2: \frac{V}{t_2} \propto \frac{1}{\sqrt{M_{\mathrm{N}_2}}}\)

For \(\mathrm{O}_2: \frac{V}{t_3} \propto \frac{1}{\sqrt{M_{\mathrm{O}_2}}}\)

From [1] and [2] we get, t2/ty = \(\sqrt{M_{\mathrm{N}_2} / M_{\mathrm{H}_2}}\) and form and we get \(t_3 / t_2=\sqrt{M_{\mathrm{O}_2} / M_{\mathrm{N}_2}}.\)

Question 31. What will happen if the collisions ofthe gas molecules with each other are not perfectly elastic?
Answer:

In the case of inelastic collisions, the total kinetic energy of gas molecules decreases, leading to a decrease in molecular speeds. As a result, the gas molecules will gradually settle down at the bottom of the container, thereby causing the pressure ofthe gas to go on decreasing gradually. A time will come when the pressure ofthe gas will come duce to zero.

Question 32. At a given temperature, the root mean square velocities of the molecules of gases A, and B are x and ycm. s^-1, respectively. If x is greater than y, then which gas has a larger molar mass?
Answer:

⇒ \(\text { For gas } A: c_{r m s}=\sqrt{\frac{3 R T}{M_A}}=x \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

⇒ \(\text { For gas } B: c_{r m s}=\sqrt{\frac{3 R T}{M_B}}=y \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

∴ \(\frac{x}{y}=\sqrt{\frac{M_B}{M_A}} ;\) Since x> y, MB will be greater than MA.

Question 33. According to the kinetic theory of gases, the average kinetic energies of O₂ and N₂ molecules are the same at a particular temperature. State whether the velocities of the molecules ofthe two gases at a given temperature will be the same or not.
Answer:

The average kinetic energy of the molecules of a gas depends only on the absolute temperature of the gas. It does not depend on the mass of gas molecules. At a given temperature, a lighter gas molecule has the same average kinetic energy as that of a heavier gas molecule.

On the contrary, the root mean square velocity of the molecules of a gas at a given temperature is inversely proportional to the molar mass ofthe gas. Hence, at a given temperature, the root mean square velocities of the molecules of N2 and O₂ gas will not be the same.

Question 34. At a given temperature, the most probable velocity of the molecules of gas A is the same as the average velocity of the molecules of gas B. Which has a larger molar mass
Answer:

Suppose, the molar masses of gases A and B are MA And MB, and the temperature of body gases is 7′ K. Therefore, at this temperature, the most probable velocity of the molecules of gas A

⇒ \(c_m=\sqrt{\frac{2 R T}{M_A}}\) and the average velocity ofthe molecules of gas B, c

⇒  \(\bar{c}=\sqrt{\frac{8 R T}{\pi M_{\mathrm{B}}}}\) As given, cm = c.

∴ \(\sqrt{\frac{2 R T}{M_A}}=\sqrt{\frac{8 R T}{\pi M_{\mathrm{B}}}} \quad \text { or, } \frac{1}{M_A}=\frac{4}{\pi M_{\mathrm{B}}}\)

∴ \(M_B=1.27 M_A\)

Therefore, gas B has a higher molar mass

Question 35. A real gas follows the equation P(V- nb) = nRT under all conditions of temperature and pressure. Show that the compressibility factor of this gas is always greater than one.
Answer:

The equation of state for the gas is: P(V-nb) = nRT The compressibility factor of the gas can be expressed as

⇒ \(P V-P n b=n R T^{\prime} \text { or, } P V=n R T+P n b\)

Or \(\frac{P V}{n R T}=1+\frac{P b}{R T} \text { or, } Z=1+\frac{P b}{R T}\)

As \(\frac{P b}{R T}\) is always positive the value of z will be greater than 1.

Question 36. For H₂ gas: a = = 0.024 L₂ atm.mol^-2 b = 0.026 Lmol^-1 a = 2.28 L₂-atm .mol^-2, b = 0.042 L-mol^-1. and for CH₄ gas:

1. At ordinary temperature and pressure, which one ofthe two gases will behave more like an ideal gas?
2. Which one of the two gases has a larger molecular size?

Answer:

1. At ordinary temperature and pressure, a real gas behaves more like an ideal gas if the values of ‘a’ and ‘b’ are very small. The values of ‘a’ and ‘b ‘ for H₂ gas are smaller than those for CH₄ gas. Obviously, at ordinary temperature and pressure, H₂ gas will behave more like an ideal gas.
2. The value of ‘b’ for a real gas reflects the sizes of molecules of the gas. A gas whose molecules are large has a high value of ‘b’. As the value of CH₄ gas is greater than that of H₂ gas, the size of the CH₄ molecule will be larger than that of the H₂ molecule.

Question 37. When does the effect of molecular volume dominate over
Answer:

The compressibility factor of a real gas becomes Z>1 when the effect of molecular volume dominates over the effect of intermolecular forces of attraction. Again, Z will be greater than one for a real gas if the pressure ofthe gas is very high. Therefore, the effect of molecular volume becomes greater than the effect of intermolecular forces of attraction if the pressure ofthe gas is very high.

Question 38. At ordinary temperature, why can CO₂ but not O₂ gas be liquefied by applying pressure? Give reason.
Answer:

A gas can be liquefied by applying the necessary pressure if its temperature is equal to or below its critical temperature. The critical temperature of CO₂ is above the ordinary temperature (usually 25°C), while that of O₂ gas is well below one ordinary temperature. Hence, CO₂ can be liquefied by applying pressure at ordinary temperature.

Question 39. The critical temperature and the critical pressure of gas are Tc and Pc, respectively. If the gas exists at a temperature of T and a pressure of P, then under which of the following conditions will the gas not be liquefied?

• T> Tc; P>PC
• r=rc; P>PC
• T = Tc; P<PC
• T<TC-P = PC

Answer: Under the conditions of T> Tc and P> Pc, the gas cannot be liquefied because its temperature is above under the conditions of T = Tc and P>PC, it is possible to liquefy the gas. Because the gas Is at its critical temperature and its pressure is above critical pressure.

Under the conditions of T = Tc and P<PC, it is not possible to liquefy the gas as the minimum pressure needed to liquefy a gas at its critical temperature must be equal to Pc or greater than Pc.

Under the conditions of T< Tc and P = Pc, the gas can be liquefied. Because the gas is below its critical temperature and the pressure of the gas is equal to its critical pressure, the minimum pressure required to liquefy a gas at its critical temperature.

Question 40. The van dar Waab constant ‘a’ for CO₂ and CH₄ gases are 3.6 and 2.3 L²-atm-mol^-2 Which one of these two gases can easily be liquefied?
Answer:

The van der Waals constant for a real gas is a measure ofthe strength of intermolecular forces of attraction in the gas. The stronger the intermolecular forces of attraction in a gas, the greater the value. Again, a gas with stronger intermolecular forces of attraction can easily be liquefied. As the value of CO₂ is greater than that of CH₄, it will be easier to liquefy CO₂ gas.

Question 41. The values of van der Waals constants ‘a’ and ‘b’ for X, Y, and Z gases are 6, 6, 20, and 0.025, 0.15, and 0.11, respectively. Which one has the highest critical temperature?
Answer:

The greater the magnitude of intermolecular forces of attraction of a gas, the higher the critical temperature it will have. Among the given gases, the value of a is maximum for gas Z. Thus, the magnitude of effective intermolecular forces of attraction is also maximum for Z, resulting in its higher value of critical temperature.

Question 42. At 20°C, the surface tension of water is three times that of CCI₄ —give reason.
Answer:

The surface tension of a liquid depends on the magnitude of intermolecular forces of attraction. It decreases or increases, respectively, with a decrease or increase in the magnitude of intermolecular forces of attraction. The only attractive forces that exist in carbon tetrachloride are to London fores exists. As the H-bond is stronger than the London force, the surface tension of water is 3 times that of CCl₄ at 20°C

Question 43. At constant pressure, the value of V/T for different quantities of an ideal gas will be different. Is this statement true or false?
Answer:

The equation of state for n moles of ideal gas is: PV=nRT

Therefore \(\frac{V}{T}=\frac{n R}{P}\)

⇒ \(\frac{V}{T}=\frac{n R}{P}\)

When the pressure remains constant, V/ T ∝ n [since R= constant]

Thus, at constant pressure, V/T for different quantities of an ideal gas will be different.

Question 44. A closed container holds a mixture of H₂, SO₂ and CH₄ gases, each with an amount of 0.5 mol. If these gases effuse through a fine orifice in the container, arrange them in the increasing order of their partial pressures once the effusion begins.
Answer:

According to Graham’s law of diffusion, at constant temperature and pressure, the rate of effusion of a gas is inversely proportional to the square root of its molar mass, i.e \(r \propto \frac{1}{\sqrt{M}}.\)

The increasing order of molar masses ofthe given gases is H₂ < MCH₄ < MSO₂– SO₂, and the order of the rate of effusion of these gases at a particular temperature and pressure will be rH₂>rCH₄>rSO₂– Once effusion begins, the order of their number of moles will be nu2 < raCH4< WSO₄ Therefore, the order of their partial pressures will be pH <pCH <PSO₂.

Question 45. The value of the compressibility factor (Z) for a gas at STP is less than 1. What is the molar volume of this gas at STP?
Answer:

We know, \(Z=\frac{V}{V_i}\) = molar volume of a real gas at a certain temperature arid pressure, V₂ – molar volume of an ideal gas at the same temperature and pressure.]

According to the question, Z <1

∴ \(\frac{V}{V_i}<1 \text { or, } V<V_i \text { or, } V<22.4 \mathrm{~L}\)

Since the molar volume of an ideal gas at STP = 22.4 litres] The volume ofthe real gas at STP will be less than 22.4 litres.

Question 46. At 0°C, plots of PV vs P for three real gases A, B and C are given below.

1. Which gas is present above its Boyle temperature? 
2. Which gas can be liquefied more easily?

Answer:

1. The PV vs P plot for gas above its critical temperature does not possess any minimum, and the value of PV increases continuously with pressure from the beginning. So, gas A is present above its Boyle temperature.
2. According to the given figure, the depth of minimum in the PV vs P curve is maximum for the gas C. Hence, the compressibility of gas C is greater than that of either A or B gas. This implies that the forces of attraction between the molecules are stronger in gas C than other two gases. Again, the stronger the intermolecular forces of attraction in a gas, the easier it is to liquefy the gas. So, gas C can be liquefied more easily.

Question 47. Rubber balloon filled with H₂ gas gets deflated after some time—explain why.
Answer:

As rubber is a porous substance, a rubber balloon contains many invisible pores on its surface. The balloon contains H₂ gas at high pressure. But the pressure of air outside the balloon is comparatively lower.

As a result, H₂ gas escapes from the balloon by effusion. This is why the pressure inside the balloon gradually falls and after some time the balloon gets deflated.

Question 48. For a real gas which obeys the van der Waals equation, a graph is obtained by plotting the values of P Vm along the y-axis and the values of P along the x-axis. What is the value of the intercept on the y-axis of the graph?
Answer:

The plot of JPVM against P may give a graph like A or B. In both cases, the graphs intersect the y-axis at P = 0. Now at very low pressure (i.e., P), the van der Waals equation reduces to the ideal gas equation.

∴ PV = nRT

or \(P\left(\frac{V}{n}\right)=R T\)

or, PVm= RT, [Vm= molar volume]

So, the value intercept on the y-axis is RT.

Question 49. The molecular speeds of gas molecules are analogous to the speeds of rifle bullets. Why is the odour of a gas not detected so fast?
Answer:

Gas molecules move almost at the same speed as rifle bullets, but the molecules do not follow a straight line path. Since the molecules collide with each other at a very fast rate, the path becomes zig-zag. Hence, the odour of a gas can not be detected as fast as its molecules move.

Question 50. Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a balloon to a certain volume at 100°C if 3.5g He is required to inflate the balloon to half the volume at 25°C.
Answer:

Volume of 3.5gHe at 25°C and pressure P is \(V=\frac{n R T}{P}=\frac{3.5}{4} \times \frac{R T}{P}=\frac{3.5 \times 298 R}{4 P}\)

To fill the balloon with H₂ at 373 K, the volume of Hg gas required, VH = 2V.

Hence \(2 V=\frac{n R T}{P}=\frac{w}{2} \times \frac{R \times 373}{P}\)

Dividing [2] by [1] gives \(2=\frac{w \times 373}{2} \times \frac{4}{3.5 \times 298} \quad \text { or, } w=2.796 \mathrm{~g}\)

Question 51. The given figure indicates the plot of vapour pressure vs. temperature for the three liquids, A, B & C. Arrange them in the increasing order of their intermolecular forces of attraction and normal boiling points.
Answer:

According to the given plot, at a particular temperature, the vapour pressure of A is higher than that of B, which in turn is higher than C. Now, a liquid with weak intermolecular forces of attraction has high vapour pressure. Therefore, the order of intermolecular forces of attraction of the given liquids will be A < B < C.

The lower the vapour pressure of a liquid, the higher its normal boiling point. Alternatively, the higher the vapour pressure of a liquid, the lower its normal boiling point. Therefore, the order of normal boiling points of the given liquids will be: A < B < C

Question 52. Water spreads on a glass surface but It forms beads on a glass surface polished by paraffin—why?
Answer:

Adhesive forces between the molecules of glass and water are stronger than the cohesive forces between the water molecules. For this reason, water can spread on the glass surface.

On the other hand, adhesive forces between the molecules of paraffin (non-polar) and water are weaker than cohesive forces between the water molecules. For this reason, water forms small beads on a glass surface of polished paraffin.

Question 53. 4 gas-mixture consists of two gases, A and B, each with equal mass. The molar mass of B is greater than that of 4. Which one of the two gases will contribute more to the total pressure of the gas mixture?
Answer:

In a gas mixture, the component gas with higher partial pressure will have a greater contribution to the total pressure of the mixture. Masses of A and B in the mixture are the same but the molar mass of B is greater than that of A. Hence, in the mixture, the number of moles or the mole fraction (xA) of A will be greater than that (xB) of B. Suppose, in the mixture, the partial pressures of A and B are pA and pB respectively and the total pressure is P. According to Dalton’s law of partial pressures, pA = xAP and pB = xgP.

As xA>xB, pA will be greater than pB. Hence, the contribution of gas A to the total pressure of the mixture will be more than that of gas B.

Question 54. Under the same conditions of temperature and pressure, the rate of diffusion of hydrogen gas is four times that of oxygen gas—explain
Answer:

At constant temperature and pressure, rates of diffusion (r) of different gases are inversely proportional to the square roots of their molecular masses (M) \(r \propto \frac{1}{\sqrt{M}}\)

⇒ \(\frac{r_{\mathrm{H}_2}}{r_{\mathrm{O}_2}}=\frac{\sqrt{M_{\mathrm{O}_2}}}{\sqrt{M_{\mathrm{H}_2}}}=\frac{\sqrt{32}}{\sqrt{2}}=4 \text { or, } r_{\mathrm{H}_2}=4 \times r_{\mathrm{O}_2}\)

So, under the same conditions of temperature and pressure, the rate of diffusion of H2 gas is four times that of O₂ gas.

Question 55. Four tyres of a motor car were filled with nitrogen, hydrogen, helium and air. In which order are these tyres to be filled with the respective gases again
Answer:

According to Graham’s law of diffusion, under the conditions of temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

The order of molar masses of the given gases is:

⇒ \(M_{\mathrm{H}_2}<M_{\mathrm{He}}<M_{\mathrm{N}_2}<M_{\text {air }}\)

At the same temperature and pressure, the order of rates of diffusion of these gases would be—rH > rHe > rN > air After a certain time, the order of decrease in pressure in these three tyres tyre (air) < tyre (N₂) < tyre (He) < tyre (H₂). Hence, tyres are to be filled again with the respective gases in the following order, tyre(H₂) tyre(He), tyre(N₂), and tyre(air).

Question 56. When a football is pumped, both the volume and pressure of the gas inside it increase. Does this observation contradict Boyle’s law?
Answer:

According to Boyle’s law, at a constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure. When the football is pumped, the quantity of air inside the football goes on increasing. As a result, the mass of air inside the football does not remain constant. Moreover, 2 pumping causes a rise in the temperature of air inside the football. Thus, neither the temperature nor the mass of the air remains constant. So, Boyle’s law is not applicable in this case.

Question 57. The molar mass of UF₆ is 176 times as high as that of H₂, yet at a particular temperature, the average kinetic energy of both is found to be the same—why?
Answer:

According to the kinetic theory of gas, the average kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas, and it does not depend upon the molar mass of the gas.

The average kinetic energy of a molecule of a gas at temperature TK is given by

⇒  \(\frac{3}{2} k T\) is Boltzmann constant].

Since the value of is constant at a given temperature and is independent of the mass of the gas molecule,

⇒   \(\frac{3}{2} k T\)

The average kinetic energy of the molecule of a heavier gas will be the same as that of the molecule of a lighter gas. Thus, at a given temperature, the average kinetic energy of a UF₆ molecule will be the same as that of a H₂ molecule.

Question 58. Two gases, obeying the van der Waals equation, have identical values of ‘b’ but different values of ‘a’. Which one of the two gases will occupy less volume under identical conditions? If the values of ‘a’ for the two gases are the same but the values of‘V are different, then under identical conditions which gas will be more compressible?
Answer:

The larger the value of ’a’ of a gas, the stronger the intermolecular forces of attraction in the gas. So, under identical conditions, a gas with a larger value of ‘a’ will be more compressible than that with a smaller value. Hence, between the two gases, the one with a higher value of ‘a’ will occupy less volume under identical conditions.

If the value of ‘a’ for two gases is the same the values of ‘b’ differ, then the gas with a smaller value of ‘b’ will be more compressible because a small value of ‘b’ for a gas signifies that the volume occupied by the molecules of the gas is small. So, this gas can be compressed to a greater extent.

Question 59. Under what conditions can a gas be liquefied?

• T = Tc and P < Pc
• T<TC and P – Pc

Answer:

It is possible to liquefy a gas by the application of pressure, provided that the temperature of the gas is equal to or less than its critical temperature. If the temperature (T) of a gas is equal to its critical temperature ( Tc), then it is possible to liquefy the gas if the pressure (P) of the gas is equal to or above its critical pressure (Pc).

So, under the condition, the gas cannot be condensed into a liquid. On the other hand, if the temperature (T) of a gas is below its critical temperature (Tc), then the gas can be transformed into liquid if the applied pressure on the gas is greater than or less than or equal to its critical pressure. So, under certain conditions, the gas can be liquefied.

Question 60. Derive the van der Waals equation for ‘n’ mol of a real gas from the equation for 1 mol of the gas.
Answer:

Van der Waals equation for 1 mol of real gas is given by:

⇒ \(\left(P+\frac{a}{v^2}\right)(\nu-b)=R T\) m……………………….(1)

If at a pressure P and a temperature T, the volume of ‘n’ moles of this gas is V, then

v = \(\frac{V}{n}\) Putting  v = \(\frac{V}{n}\) in equation n n [1], we have

⇒ \(\left(P+\frac{a}{\left(\frac{V}{n}\right)^2}\right)\left(\frac{V}{n}-b\right)=R T \text { or }\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T\)

Question 61. Between methanol (CH₃OH) and water (H₂O) whose surface tension is greater, and why?
Answer:

Molecules in methanol and water both are capable of forming hydrogen bonds. However, the number of hydrogen bonds per molecule in water is greater than that in methanol. So, the intermolecular forces of attraction are stronger in water than those in methanol. Again, the stronger the intermolecular forces of attraction in a liquid, the greater the surface tension of the liquid. So, the surface tension of water is greater than that of methanol.

Question 62. A liquid has a high normal boiling point. Will its viscosity and surface tension values be high or low?
Answer:

The high normal boiling point of a liquid indicates that the liquid possesses strong intermolecular forces of attraction. The values of viscosity and surface tension of a liquid depend upon the strength of intermolecular attractive forces in the liquid. The stronger the intermolecular attractive forces, the higher the values of viscosity and surface tension of the liquid. As the intermolecular forces of attraction are strong for the concerned liquid, the values of both surface tension and viscosity of the liquid will be high.

Question 63. How is the vapour pressure of a liquid affected if the surface area of the liquid is increased at a given temperature?
Answer:

• At a given temperature, the vapour pressure of a liquid does not depend upon the surface area of the liquid. If the surface area of a liquid is increased, keeping the temperature constant, then its rate of evaporation increases because at a particular time, more molecules are now able to leave the liquid surface and go to the vapour phase.
• The increased surface area of the liquid also causes the increase in the die rate of condensation because, at a particular time, more molecules are now able to reenter from vapour to the liquid phase.
• Therefore, the rates of both evaporation and condensation are increased to the same extent So, the vapour pressure remains the same at a constant temperature and is not affected by the surface area of the liquid

Question 64. Compare the viscosity coefficients of the following liquids at a particular temperature: Propanol (CH₃CH₂CH₂OH), ethylene glycol (HOCH₂—CH₂OH) and glycerol (HOCH₂—CHOH—CH₂OH).
Answer:

The number of OH groups in the molecules of CH₃CH₂CH₂OH, HOCH,—CH₂OH HOCH₂—CHOH—CH₂OH are 1, 2 and 3, respectively. So, the number of hydrogen bonds formed per molecule of CH₃CH₂CH₂OH HOCH₂—CH₂OH and HOCH₂ —CHOH —CH₂OH are 1, 2 and 3, respectively. Thus, the order of the intermolecular forces of attraction of the given liquids will be propanol < ethylene glycol < glycerol.

Again, the stronger the intermolecular forces of attraction of a liquid, the higher the value of its viscosity coefficient. Thus, the order of viscosity is propanol < ethylene glycol < glycerol.

Question 65. Find out the minimum pressure required to compress 2× M 500 dm³ of air at bar to 200 dm1 at 30nC
Answer:

In this process, the mass and temperature of the gas remain constant but the volume and pressure of the gas change.

Given P₁=1 bar, V₁= 500 dm³, v₂= 200 dm³.

Appyling Boyle’s law , pe \(P_2=\frac{P_1 V_1}{V_2}=\frac{1 \times 500}{200}=2.5 \text { bar }\)

So, the minimum pressure needed to V₂ compress 200 the gas is 2.5 bar.

Question 66. A vessel of 120 mi. capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 1 HO ml. at 35°C. What would be Its pressure?
Answer:

In this process, the pressure and volume ofthe gas change but its mass and temperature remain constant.

Applying Boyle’s law to the process, we have \(P_2=\frac{P_1 V_1}{V_2}\)

Given: p₁= 1.2 bar, V₁ = 120ml and V₂ = 180 ml

∴ \(p_2=\frac{1.2 \times 120}{180}=0.8 \text { bar }\)

Question 67. A real gas follows van der Waals equation. Find the compressibility for 1 mol of the gas at its critical temperature
Answer:

At the critical point

⇒ \(V=V_c=3 b ; P=P_c=\frac{a}{27 b^2} \text { and } T=T_c=\frac{8 a}{27 R b}\)

∴ \(Z=\frac{P V}{R T}=\frac{\frac{a}{27 b^2} \times 3 b}{R \times \frac{8 a}{27 R b}}=\frac{3}{8}\)

NCERT Solutions For Class 11 Chemistry Chapter 5 States Of Matter Gases And Liquids Warm-Up Exercise Question And Answers

Question 1. Under which of the following conditions will the density of a, fixed mass of SO₂ gas be higher?— STP 27°C and 3atm pressure.
Answer:

d = \(\frac{P M}{R T}\) and M is constant for a particular gas. Hence ∝PIT. The value of PIT at 3atm and 27°C is greater than that at STP.

Therefore, the density of SO₂ at 3atm and 27°C will be greater than that at STP.

Question 2. Determine the SI unit of
Answer:

⇒ \(\frac{P V^2 T^2}{n}=\frac{\mathrm{N} \times\left(\mathrm{m}^3\right)^2 \times \mathrm{K}^2}{\mathrm{~m}^2 \times \mathrm{mol}}=\mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{K}^2 \cdot \mathrm{mol}^{-1}\)

Question 3. A gas mixture consists of three gases A, B, and C with the number of moles 1, 2, and 4, respectively. Which of these gases will have a maximum partial pressure if the total pressure ofthe mixture is Pata given temperature t what temperature will the average velocity of O₂ molecules be equal to that of U₂ molecules at 20K?
Answer:

In the mixture, the total number of moles of the constituent gases =(1 + 2 + 4) = 7mol.

The mole fractions of A, B, and C

⇒  \(\frac{1}{7}, \frac{2}{7} \text { and } \frac{4}{7} \text {, }\) respectively.

Since the mole fraction of C is the highest, its partial pressure will be the highest in the mixture.

Question 4. Rank Cl₂, SO₂, CO₂, and CH₄ gases in Increasing order of their rates of diffusion under identical set of conditions.
Answer:

The molar mass of the gases follows the order: \(M_{\mathrm{CH}_4}<M_{\mathrm{CO}_2}<M_{\mathrm{SO}_2}<M_{\mathrm{Cl}_2}\)

Hence, at constant temperature and pressure, the rates of diffusion will be the order: \(r_{\mathrm{Cl}_2}<r_{\mathrm{SO}_2}<r_{\mathrm{CO}_2}<r_{\mathrm{CH}_4}\)

Question 5. A closed vessel holds a gas mixture consisting of C₂H₆, C₂H₄, and CH₄, each with an amount of 2.5 mol. However, due to a pinhole in the vessel, the gas mixture undergoes effusion. What will be the order of partial pressures of the gases in the vessel after some time?
Answer:

Molar masses of C₂H₆, C₂H₄ and CH₄ follow the order:

⇒ \(M_{\mathrm{CH}_4}<M_{\mathrm{C}_2 \mathrm{H}_4}<M_{\mathrm{C}_2 \mathrm{H}_6}\)

Thus, at a certain temperature and pressure, their rates of effusion will be in the order \(r_{\mathrm{C}_2 \mathrm{H}_6}<r_{\mathrm{C}_2 \mathrm{H}_4}<r_{\mathrm{CH}_4}\)

Therefore, the order of partial pressure after some time will be:

⇒ \(p_{\mathrm{CH}_4}<p_{\mathrm{C}_2 \mathrm{H}_4}<p_{\mathrm{C}_2 \mathrm{H}_6}\)

Question 6. At constant temperature and pressure volume of an ideal gas (molecular mass 28 g. mol^-1) is 23.36 times greater than its mole number. Find out its density at the same temperature and pressure.
Answer:

Let us suppose, the tire volume and density of n mol of an ideal gas are V L and d g.L^-1at P atm and T K.

Now, PV = nRT and d \(=\frac{P M}{R T}\)i.e., \(d=M \times \frac{n}{V}\)

Question 7. Why cannot CO₂ gas be liquefied above 31.1°C?
Answer:

The critical temperature of CO₂ is 31.1°C. Above 31.1°C, due to the very high average kinetic energy of CO₂ molecules, the attraction between them becomes negligible. As a result, CO₂ gas cannot be liquefied above 31.1°C.

Question 8. Under what conditions will the value of- always be the same irrespective ofthe value of T?
Answer:

For a fixed mass of gas at constant volume, P ∝ T or, P = f Cx T (K = constant). Therefore, P/T = K. This relation indicates that the value of P/T is always constant for a fixed mass of a gas at constant volume irrespective of the value of T is

Question 9. Under what conditions will the value of PV always be the same irrespective of the value of P (or V)?
Answer:

According to Boyle’s law, for a fixed mass of gas at a constant temperature, PV = constant. Therefore, the value of PV is always constant for a fixed mass of a gas at a constant temperature irrespective ofthe value of P

Question 10. 1 mol of N₂ & 3 mol of O₂ are kept In two different containers with a volume of V at a fixed temperature. Compare—(1) the average kinetic energy and (11) the total kinetic energy ofthe molecules.
Answer:

The average kinetic energy of gas molecules depends only on the absolute temperature ofthe gas. Since both gases are at the same temperature, they will have equal average kinetic energy.

At T K, the total kinetic energy of nmol gas molecules

⇒ \(n \times \frac{3}{2} R T\) Hence, at T K, the total kinetic energy of 3mol O₂ molecules is 3 times that of 1 mol N₂ molecules.

Question 11. For a real gas, the van der Waals constant ‘a’ is zero. Can the gas be liquefied? Explain.
Answer:

The van der Waals constant ‘a’ measures the magnitude of intermolecular forces of attraction in a gas. Hence, a real gas with ‘a’= 0 signifies that there are no intermolecular attractive forces in the gas. Consequently, such a gas cannot be liquefied.

Question 12. State Gay Lussac’s law related to the pressure and temperature of a gas. 3.2 g of sulphur when vaporised, the sulphur vapour occupies a volume of 280.2 mL at STP. Determine the molecular formula of sulphur vapour under this condition. (S = 32)
Answer:

280.2 mL of sulphur weighs 3.2 g at STP. 22400 mL of sulphur weighs 255.8 g at STP. Let, the molecular formula of sulphur Sn. So, n × 32 = 255.8 or, n = 7.9 ~ 8 Molecular formula ofsulphur is Sg.

Question 13. At a constant temperature, a container of fixed volume holds NH3 and HCl gases. Can Dalton’s law of partial pressures be applied to this gas mixture?
Answer:

Dalton’s law of partial pressure applies only to a mixture composed of two or more non-reacting gases. NH₃ and HCl gases react together to produce NH₄Cl. So Dalton’s law of partial pressure will not be applicable in this case.

Question 14. At tC and t2°C, the values of viscosity coefficients of a liquid are x poise, and y poise respectively. If x>y, then which one is higher, t₁ or t₂?
Answer:

With the increase in temperature, the viscosity of a liquid decreases. Now, viscosity directly varies with the value of viscosity coefficients As the viscosity coefficient of the liquid at t2°C is smaller than that at t₁°C, t₂ > t₁.

Question 15. The equation of state of a real gas is P(V-b) = RT. Can the gas be liquified? Explain. Sketchlog P vlog V graph for a given mass of an ideal gas at constant temperature and indicate the slope.
Answer:

Value of van der Waals constant V= 0 for the given gas. Hence, there exists no force of attraction among the gas molecules. So, the gas cannot be liquefied.

Question 16. Determine the types of intermodular forces of attraction in the following instances. w-hexane, SO₂,CO₂, CHCI₃, (CH₃)₂CO, (CH₃)₂O
Answer:

London forces or instantaneous induced dipole instantaneous induced dipole attraction: n-hexane, CO₂, Dipole-dipole attraction: SO₂ CHCI₃, (CH₃)₂O. However, the London force also acts in this case.

Question 17. Which type of intermodular forces of attraction act between O₂ and water modules when O₂ is dissolved in water?
Answer:

O₂ is a non-polar molecule, whereas the H₂O molecule is polar. Hence, the force of attraction acting between O₂ and H₂O molecules is dipole-induced dipole attraction

Question 18. The critical temperatures of H₂, NH₃, and CO₂ gases are 5K, 405K, and 304K, respectively. Arrange them in the increasing order of their intermolecular forces of attraction.
Answer:

A gas with high critical temperature possesses strong intermolecular forces of attraction. The order of critical temperatures of the gases H2, NH₃, and CO₂ is NH₃ > CO₂ > H₂. So, the increasing order of their strength of intermolecular forces will be H₂<CO₂<NH₃.

Question 19. The critical temperatures of NH₃ and SO₂ gases are 405.0K and 430.3K, respectively. For which gas is the value of a der Waals constant greater, and why?
Answer:

The higher the critical temperature of a gas, the stronger its intermolecular forces of attraction, and hence the larger the value of the van der Waals constant the gas has. Thus, between NH₃ and SO₂, the value of a will be larger for SO₂ because its critical temperature is higher than that of NH₃.

Question 20. The critical temperatures of NH₃, CO₂, and O₂ gases are 405.6K, 304.1K, and 154.2K, respectively. If the gases are cooled from 500K to their respective critical temperatures, then which gas will be liquefied first?
Answer:

If the given gases are cooled from 500K, NH₃ gets liquefied first (critical temperature 405.6K). The reason is that the critical temperatures of CO₂ and O₂ are lower than that of NH₃. As a result, the liquefaction of either CO₂ or O₂ is not possible at the critical temperature of NH₃.

Question 21. At 20°C the increasing order of viscosity of acetic acid, acetone, and methanol is: acetone < methanol < acetic acid. Arrange the liquids according to their increasing intermolecular attractive forces.
Answer:

With the increase or decrease in the intermolecular force of attraction of a liquid, the value of viscosity confidence of the liquid increases or decreases. At 20°C, the increasing order of the values of the viscosity coefficient of the given liquids is acetone < methanol < acetic add. Thus, the order of increasing the intermolecular attractive force of these liquids is acetone < methanol < acetic add.

Question 22. Why does the surface energy increase on the dispersion of a liquid? large water drop into smaller droplets?
Answer:

When a large water drop disperses into smaller water droplets, the total surface area ofthe small water droplets becomes greater than the surface area of the large water drop. As the surface energy increases with an increase in surface area, the dispersion of a large water drop into smaller droplets leads to an increase in surface.

Question 23. \(c_{r m s}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 P V}{M}} .\) According to this equation, with the increase in pressure or volume, the value of arms increases. Justify this statement.
Answer:

The rms velocity of the molecules of a gas depends only on the temperature and molar mass of the gas. Its value does not depend on the pressure or volume of the gas because, at a constant temperature, PV for a given mass of gas always remains constant, irrespective of the values of pressure or volume.

Question 24. The volumes, energy numbers of the two ideal molecules gases A and and B average the same. What is the relationship between the pressures of these two gases?
Answer:

The average kinetic energy of gas molecules depends only on the temperature ofthe gas. As the molecules of A and B gases have the same values of average kinetic energy, both the gases have die same absolute temperature.

Since the volume, the number of molecules and the temperature of both gases are identical, the pressure will be the same.

Question 25. Any real gas behaves ideally at very low pressure and high temperature. Explain. The values of van der Waals constant ‘ a’ for N₂ and NH₃ are 1.37 and 4.30 L² -atm- mol¯² respectively. Explain the difference in values.
Answer:

Van der Waals constant ‘a’ denotes the magnitude of attractive forces between the molecules of real gases. The value of ‘a’ is higher in the case of NH₃ than N₂. That means attractive forces (London force, dipole-dipole attraction force) present among the molecules of NH₃ are stronger than those of N₂ (London force). Consequently, NH₃ can be liquefied more easily than N₂.

Question 26. Calculate the temperature of 4.0 mol of gas occupying 5 dm³ at 3.32 bar. (R = 0.083 bar. dm³– K^-1. mol^-1)
Answer:

To calculate the temperature ofthe gas, we apply the gas equation, PV = nRT. As given: P = 3.32 bar, V = 5 dm3 and n = 4 mol Putting these values into the equation, PV = nRT, we have

⇒ \(T=\frac{P V}{n R}=\frac{3.32 \times 5}{4 \times 0.083} \mathrm{~K}=50 \mathrm{~K}\)

Question 27. Calculate the total number of electrons present In 1.4 g of dinitrogen gas.
Answer:

⇒ \(1.4 \mathrm{~g} \text { of } \mathrm{N}_2=\frac{1.4}{28}=0.05 \mathrm{~mol} \text { of } \mathrm{N}_2\)

1 molecule of N₂ contains 14 electrons, Hence, 0.05 mol of N2, i.e., 0.05 × 0.022 × 10²³ molecules of = 4,2154 × 10²³ electrons. N2 contain 14 × 0,05 × 0,023 × 10²³

Question 28. When does the graph showing variation ofthe volume ofa given mass of gas with pressure at a constant temperature become linear?
Answer:

For a given mass of gas at a fixed temperature, PV = K (constant). That is, \(P=\frac{K}{V}\) This relation expresses an equation of a straight line passing through the origin. Therefore, if P is plotted against \(\frac{1}{V}\) for a given mass of a gas at a fixed temperature, a straight line will be obtained

Question 29. N₂ gas is present in a 1L desiccator at latm pressure. The pressure ofthe gas decreases to 78mmHg pressure when the desiccator is partially evacuated using a vacuum pump at a constant temperature. Find out the final volume of the gas.
Answer:

Since the volume of the desiccator is fixed, the final volume of the gas will be 1L even after the desiccator is partially evacuated. In this process, the number of moles of the gas decreases but its volume and temperature remain the same. As the pressure ofthe gas is reduced

Question 30. According to Boyle’s law, at a constant temperature, the volume ofa given mass of gas is inversely proportional to its pressure. But when a balloon is filled with air, both the volume and pressure ofthe gas inside it increase—Explain.
Answer:

When the balloon is pumped, the quantity of air inside the balloon goes on increasing. As a result, the mass of air inside it does not remain constant. Moreover, pumping causes a rise in the temperature of the air inside the balloon. Thus, neither the temperature nor the mass of the air remains constant. So, Boyle’s law is not applicable in this case.