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

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

Question 1. At a given pressure, the volume of a given amount of gas at 0°C is V0. Will the V vs t (celsius temperature) plot for this gas be linear? Will it be a straight line passing through the origin? If this straight line does not pass through the origin, then what will be its slope and intercept?
Answer:

According to Charles’ law, Vt = V₀ \(\left(1+\frac{t}{273}\right)\); where Vt and V₀ are the volumes of a given mass of gas at a temperature t°C respectively, when the pressure of the gas is kept constant. Hence, the above equation can be rewritten as

⇒  \(V_t=V_0+\frac{t}{273} V_0\)……………[1]

V₀ is a fixed quantity for a given mass of gas at a constant pressure. Thus, the equation [1] represents a straight-line equation. Hence, the Vt vs t plot will give a straight line.

Equation [1] does not represent an equation of a straight line passing through the origin. Hence, the Vf vs t plot will not be a straight line passing through the origin

According to the equation [1], the straight line resulting from the plot of V( vs t has a slope of V₀/273 and an intercept of V₀.

Question 2. A certain amount of an ideal gas is enclosed in a cylinder fitted with a movable piston. What would be the changes in the volume ofthe gain in the following processes?

1. The pressure of the gas is reduced by 25% at constant temperature.
2. The temperature of the gas is increased by 50% at constant pressure.

Answer:

1. According to Boyle’s law, P₁V₁ = P₂V₂ when the mass and temperature of a gas are constant. After the reduction of initial pressure (P₁) by 25%, the final pressure (P₂) becomes

⇒ \(\left(P_1-P_1 \times \frac{25}{100}\right)=0.75 P_1\)

i.e., P₂ = 0.75P₁

∴ \(V_2=\frac{P_1 V_1}{P_2}=\frac{P_1 \times V_1}{0.75 P_1}=\frac{4}{3} V_1\)

Therefore, the change in volume \(=\frac{4}{3} V_1-V_1=\frac{V_1}{3}\)

2. According to Charles’ law, V₁T₂ = V₂T₁ when the mass and pressure of a gas remain constant. After increasing the initial temperature (T₁) by 50%, the final temperature (T₂) becomes

⇒  \(\left(T_1+T_1 \times \frac{50}{100}\right)=1.5 T_1 \text {, i.e., } T_2=1.5 T_1\)

∴ \(V_1 \times 1.5 T_2=V_2 T_2 \quad \text { or, } V_2=1.5 \times V_1\)

∴ The change in volume = 1.5 V₁– V₁ = 0.5V₁

Question 3. Determine the values of molar gas constant in the following units— mL torr K^-1mol^-1; kPa-L-K^-1-mol^-1.
Answer:

R =0.0821 L-atm -K^-1. mol^-1

= 0.0821 × 103 × 760 torr -mL .K^-1.mol^-1

[Since 1L = 10³mL and 1 atm = 760

= 6.23 ×  10⁴ torr -mL -K^-1.mol^-1

= 0.0821 L .atm -K^-1.mol^-1

= 0.0821 × 1.013 x 1 02 kPa . L K^-1. mol^-1

= 8.31 kPa-L-K^-1-mol^-1

[Since 1 atm = 1.013 ×10⁵ Pa = 1.013 × 10²kPa

Question 4. For the same mass of two ideal gases X and Y at the same temperature and pressure, the volume of Y is found to be three times as large as that of X. Compare the values of their molar masses.
Answer:

Px = Py, Tx = Ty, 3VX = Vy

According to the ideal gas equation, Pxvx = nXRTx and = P_x V_x = n_xRT_x

i.e., \(\frac{P_X}{P_Y} \times \frac{V_X}{V_Y}=\frac{n_X}{n_Y} \times \frac{T_X}{T_Y} \quad \text { or, } \frac{1}{3}=\frac{n_X}{n_Y}\)

Let, molar masses of X and Y be My and My respectively

∴ \(\frac{w / M_X}{w / M_Y}=\frac{1}{3} \quad \text { or, } \frac{M_X}{M_Y}=3\)

Question 5. When a flask of fixed volume is filled with – mol of an ideal gas A at a constant temperature, the pressure ofthe gas becomes 2 atoms. Adding 2y mol of another ideal gas B to the flask at the same temperature causes the pressure of the system to increase to 4.0 atm.
Answer:

For tyre gas. A: P = 2 atm and \(n=\frac{x}{2} \mathrm{~mol}\)

Hence, PV \(=n R T \text { or, } 2 \times V=\frac{x R T}{2}\)

For gas mixture: P=4atm \(\& n=\frac{x}{2}+2 y=\frac{1}{2}(x+4 y) \mathrm{mol}\)

∴ In case of gas mixture \(4 \times V=\frac{1}{2}(x+4 y) R T\)

Deriding [2] -r [1] we get \(\frac{4}{2}=\frac{x+4 y}{x} \text { or, } x=4 y\)

Question 6. At 27°C and 1 atm pressure, the volume of a 5.0g mixture of He and Ar gases is 10dm3. Find the mass per cent of the two gases in the gas mixture.
Answer:

Let, the amount of He and Ar in the mixture be ag and bg respectively. Hence, a + b = 5

⇒ \(\text { Now, } P V=\left(n_1+n_2\right) R T \text { or, } 1 \times 10=\left[\frac{a}{4}+\frac{b}{40}\right] \times 0.0821 \times 300\)

∴ 10 a +b = 16.24

By solving [1] and [2] we get, a = 1.25g and b = 3.75g

∴ \(\% \mathrm{He}=\frac{1.25}{5} \times 100=25 \text { and } \% \mathrm{Ar}=\frac{3.75}{5} \times 100=75\)

Question 7. A gas mixture consisting of O₂ and N₂ gases has a volume of 5 L at 25°C. In the mixture, if the mass of O₂ gas is twice that of N₂ gas, then which one of them will have a greater contribution to the total pressure ofthe mixture?
Answer:

In the mixture \(n_{\mathrm{O}_2}=\frac{2 w}{32}\)

=\(\frac{w}{16} \mathrm{~mol} \text { and } n_{\mathrm{N}_2}\)

=\(\frac{w}{28} \mathrm{~mol} \text {. }\)

∴ Total number of moles (n) \(=\frac{w}{16}+\frac{w}{28}=\frac{11 w}{112} \mathrm{~mol}\)

⇒ \(\text { Hence, } x_{\mathrm{N}_2}\)

= \(\frac{w}{28} \times \frac{112}{11 w}\)

= \(\frac{4}{11} \text { and } x_{\mathrm{O}_2}\)

= \(\frac{w}{16} \times \frac{112}{11 w}=\frac{7}{11}\)

Thus, xO₂ > xN₂. So, pO₂> PN₂

(Since pi = zip)

Question 8. In a gas mixture of H₂ and He, the partial pressure of H₂ is half that of He. Find the mole fractions of H₂ and He in the mixture.
Answer:

As given, p₂ = pH₂ or, pHe = 2 × PH₂

or, xHe × P = 2 × XH₂ × P or, xHe = 2 × xe = 2 × xH₂

Hence, 2 × xH₂+ xH₂ = 1

Or, \(x_{\mathrm{H}_2}=\frac{1}{3} \approx 0.34\)

∴ xHe = 2 × XH₂ = 0.68

Question 9. A closed vessel contains an equal mass of O₂ and CH₄ gases at 25°C. What fraction of the total pressure is contributed by CH₄ gas?
Answer:

⇒ \(n_{\mathrm{CH}_4}=\frac{w}{16}, n_{\mathrm{O}_2}=\frac{w}{32}\)

∴ Total number of moles = \(\frac{w}{16}+\frac{w}{32}=\frac{3 w}{32}\)

∴ \(x_{\mathrm{CH}_4}=\frac{w}{16} \times \frac{32}{3 w}=\frac{2}{3}\)

⇒ \(p_{\mathrm{CH}_4}=\frac{2}{3}\); Thus Memthane Contributes \(\frac{2}{3}\) rd of the total pressure.

Question 10. A mixture of O₂ and H₂ gases contains 20% of H₂ gas. At a certain temperature, the total pressure of the mixture is found to be 1 bar. What is the partial pressure of O₂ (in bar) in the mixture?
Answer:

In the mixture, 20% H₂ is present. Hence extent of oxygen = 80%

∴ \(n_{\mathrm{H}_2}=\frac{20}{2}=10 \text { and } n_{\mathrm{O}_2}=\frac{80}{32}=\frac{5}{2}\)

∴ Total number of moles \(=\left(10+\frac{5}{2}\right)=\frac{25}{2}\)

∴ \(p_{\mathrm{O}_2}=\frac{5}{2} \times \frac{2}{25} \times 1 \mathrm{bar}=0.2 \mathrm{bar}\)

Question 11. Under the same conditions, a gas diffuses times as fast as an SO₂ gas. Find the molecular mass ofthe gas.
Answer:

⇒ \(\frac{r_{\text {gas }}}{r_{\mathrm{SO}_2}}=\sqrt{\frac{M_{\mathrm{SO}_2}}{M_{\text {gas }}}}=\sqrt{2}\) \(\text { or, } \frac{M_{\mathrm{SO}_2}}{M_{\text {gas }}}=2 \text { or, } M_{\mathrm{gas}}=32 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

or, 8n-1 = 15 or, n = 2

Question 12. At what temperatures rms velocity, average velocity & most probable velocity of O₂ molecules will be 1500 m.s^-1?
Answer:

Molar mass (M) of O₂ – 32g.mol^-1 =0.032kg. mol^-1

In case of rms velocity:

⇒  \(\frac{3 R T_1}{M}=(1500)^2\)

⇒ \(\text { or, } \frac{3 \times 8.314 \times T_1}{0.032}=(1500)^2 \text { or, } T_1=2886.697 \mathrm{~K} \text {. }\)

In case of average velocity:

⇒ \(\frac{8 R T_2}{\pi M}=(1500)^2\)

⇒ \(\text { or, } \frac{8 \times 8.314 \times T_2}{0.032 \times 3.14}=(1500)^2 \text { or, } T_2=3399.085 \mathrm{~K} \text {. }\)

In case of most probable velocity: \(\frac{2 R T_3}{M}=(1500)^2\)

⇒ \(\text { or, } \frac{2 \times 8.314 \times T_3}{0.032}=(1500)^2 \text { or, } T_3=4330.045 \mathrm{~K} \text {. }\)

Question 13. At a given temperature and pressure, 1 mol of an ideal gas occupies a volume of 20.8 L. For mol of a real gas at the same temperature and pressure—

• Z will be equal to 1 if the volume ofthe real gas is…
  
• Z will be greater than 1 if the volume ofthe real gas is…
  
• Z will be less than 1 if the volume ofthe real gas is…

Answer: \(Z=\frac{V}{V_i}\); the volume of a certain amount of ideal gas at a given temperature and pressure and V = the volume of the same amount of a real gas at the same temperature and pressure.

As given, V- = 20.8L. Therefore, if

• V = 20.8L, then Z will be equal to1
• V> 20.8L, then Z will be greater than1
• V< 20.8L, then Z will be less than1

Question 14. Two ideal gases A and B are mixed at temperature T and pressure P. Show that \(d=\left(X_A M_A+X_B M_B\right) \frac{P}{R T}\) Id = density of the mixture, X₂ = mole fraction of A, XD = mole fraction of B, M A = Molar mass of A, Mlt = Molar mass of B]
Answer:

Suppose, the total volume of the gas mixture is V, and nA and nB are the respective number of moles of A and B in the mixture. If W A and be the masses of A and B respectively, in the mixture, then

⇒ \(n_A=\frac{w_A}{M_A} \text { and } n_B=\frac{W_B}{M_B}\)

Total mass ofA & Bin mixture = \(W_A+W_B=n_A M_A+n_B M_B\)

∴ Density Of The Mixture \(=\frac{W_A+W_B}{V}=\frac{n_A M_A+n_B M_B}{V}\) for the gas mixture

⇒ \(P V=\left(n_A+n_B\right) R T ;\)

∴ \(V=\left(n_A+n_B\right) \frac{R T}{P}\)

⇒ \(\text { So, } d=\frac{n_A M_A+n_B M_B}{n_A+n_B} \times \frac{P}{R T}=\left(\frac{n_A M_A}{n_A+n_B}+\frac{n_B M_B}{n_A+n_B}\right) \frac{P}{R T}\)

∴ \(\left.d=X_A M_A+X_B M_B\right) \frac{P}{R T}\left[X_A=\frac{n_A}{n_A+n_B} ; X_B=\frac{n_B}{n_A+n_B}\right]\)

Question 15. “The total kinetic energy of the molecules in an ideal gas with a volume V at pressure P and temperature T is equal to the total kinetic energy of the molecules present in the same volume of another ideal gas at the same pressure and temperature 2T”—Justify the statement.
Answer:

Suppose, the number of molecules present in volume V of the first gas =. If the root mean square velocity of the molecules in the first gas is cl, and the mass of each molecule is mx, then from the kinetic gas equation we have,

⇒ \(P V=\frac{1}{3} m_1 n_1 c_1^2 \quad \text { or, } P V=\frac{2}{3} n_1 \times \frac{1}{2} m_1 c_1^2\)

Or \(P V=\frac{2}{3} E_1 \text { [where } E_1=n_1 \times \frac{1}{2} m_1 c_1^2\)

Energy ofthe molecules.] Similarly, if the number of molecules in volume V of the second gas = n₂, the mass of each molecule = m2 and the root mean square velocity of molecules= c₂, then

⇒ \(P V=\frac{1}{3} m_2 n_2 c_2^2=\frac{2}{3} n_2 \times \frac{1}{2} m_2 c_2^2 \quad \text { or, } P V=\frac{2}{3} E_2\)

[where E₂ = total kinetic energy ofthe molecules of second gas] As P and V of both the gases are equal, so E₁ = E₂.

Question 16. Prove that at a certain pressure, the rate of diffusion of a gas is proportional to the square root of the I absolute temperature of the gas
Answer:

The rate of diffusion of a gas at constant pressure (r) oc root mean square velocity ofthe gas (Crms)

∴ \(r \propto c_{r m s} \text { or, } r \propto \sqrt{\frac{3 R T}{M}}\)

Since \(c_{r m s}=\sqrt{\frac{3 R T}{M}}\)

At constant pressure, if and r2 are the rates of diffusion of a particular gas at temperatures and T2, respectively, then

⇒ \(r_1 \propto \sqrt{\frac{3 R T_1}{M}} \text { and } r_2 \propto \sqrt{\frac{3 R T_2}{M}}\)

∴ \(\frac{r_1}{r_2}=\sqrt{\frac{T_1}{T_2}} \text { or, } r \propto \sqrt{T}\)

So, at constant pressure, the rate of diffusion of a gas is proportional to the square root of the absolute temperature of the gas.

Question 17. At constant temperature and pressure, the compressibility factor (Z) for one mole of a van der Waals gas is 0.5. If the volumes of the gas molecules are considered to be negligible, then show that \(a=\frac{1}{2} V_m\) where Vm and Tare the molar volume and temperature of the gas respectively.
Answer:

We known \(Z=\frac{P V}{n R T}\); Given Z= 0.5 and n=1

∴ \(P V=0.5 R T=\frac{1}{2} R T\)

The equation of state for mol of a van der Waals gas is, \(\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T\) [Vm= molar volume]

If volumes of the molecules are considered to be negligible, as per the given condition, then \(V_m-b \approx V_m\)

∴ \(\left(P+\frac{a}{V_m^2}\right) V_m=R T \quad \text { or, } P V_m+\frac{a}{V_m}=R T\)

Again, from equation [1] we get, \(P V=\frac{1}{2} R T\)

For 1 mol (n = 1) ofthe gas, V = Vm (molar volume)

Therefore \(P V_m=\frac{1}{2} R T\)

From equations [2] and [3] we have,

⇒ \(\frac{1}{2} R T+\frac{a}{V_m}=R T \quad \text { or, } a=\frac{1}{2} V_m R T\)

Question 18. At a given temperature and pressure, the volume fraction of an ideal gas is equal to its mole fraction in a mixture of ideal gases—is it true or false?
Answer:

Suppose, at a given temperature (T) and pressure (P), the volumes of two ideal gases are VA and Vg, respectively, and n_A and n_B are their respective number of moles. Let at the same temperature and pressure, the total volume ofthe mixture of these two gases is V.

∴ According to Amagat’s law of partial volume, \(V=V_A+V_B\)

So, the partial volume of A in the mixture \(=\frac{V_A}{V}\) and its mole fraction \(=\frac{n_A}{n_A+n_B}\)

Now, applying the ideal gas equation to each component gas as well as the gas mixture, we get

PV_A = n_ART ……………………(1)

PV_B=n_BRT ……………………(2)

PV = (n_A + n_B)RT ……………………(3)

Dividing equation no. [1] by equation no. [3], we have

⇒ \(\frac{P V_A}{P V}=\frac{n_A}{n_A+n_B}\)

∴ \(\frac{V_A}{V}=\frac{n_A}{n_A+n_B}\)

That, the volume fraction of A = the mole fraction of A

Similar \(\frac{V_B}{V}=\frac{n_B}{n_A+n_B}\)

[Dividing equation no. [2] by equation no. [3]

That is the volume fraction of B = the mole fraction of B. Therefore, at a given temperature and pressure, the volume fraction of an ideal gas in a mixture of ideal gases is equal to its mole fraction. Hence, the given statement is true.

Question 19. A 15.0 L vessel containing 5.6 g of N₂ is connected to a 5.0 L vessel containing 8.0 g of O₂ using a valve. After the valve is opened and the gases are allowed to mix, what will be the partial pressure of each gas in the mixture at 27°C?
Answer:

5.6g N₂ = \(\frac{5.6}{28}=0.2 \mathrm{~mol}\) N₂

8.0 g O₂ = \(\frac{8}{32}=0.25 \mathrm{~mol}\) O₂

= \(\frac{5.6}{28}=0.2 \mathrm{~mol} \mathrm{~N}_2, 8.0 \mathrm{~g} \mathrm{O}_2\)

=\(\frac{8}{32}=0.25 \mathrm{~mol} \mathrm{O}_2\) and

T = (273 + 27)K = 300 K.

After the opening of the valve, the total volume ofthe gas mixture, V = (15 + 5)L = 20 L

If the partial pressure of N₂ gas in the mixture is pN, then pN x V = nN RT or,

PN₂ × 20=0.2 × 0.0821× 300

∴ PN₂ = 0.2463 atm

If the partial pressure of O₂ gas in the mixture is pO₂

then XV= NO₂RT

Or, PO₂ × 20=0.25 × 0.00821 × 300

∴ PO₂= 0.30+748 atm

Question 20. Why is the quantity of air required to inflate the tyre of a car in summer less than that required in winter?
Answer:

• According to the kinetic theory of gases, the pressure of a gas originates due to the bombardment of the gas molecules on the walls ofthe container. At higher temperatures, the molecules in a gas have higher velocities or average kinetic energy, and they collide with the walls of their container more frequently and with greater force.
• As a result, the pressure of a gas increases with an increase in temperature if the volume of the gas remains fixed.
• The summer temperature is higher than the winter temperature, and the average kinetic energy of the air molecules in summer is comparatively more than that in the winter.
• As a result, the air molecules in the summer will exert a greater amount of force on the walls than that exerted by the same number of molecules in the winter.
• Therefore, if the volume remains fixed, the pressure of a certain amount of air in summer will be more than that in winter.

Question 21. Two flasks of equal volume are connected by a narrow tube of negligible volume and are filled with N₂ gas. When both the flasks are immersed in boiling water the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water while the other flask is in boiling water
Answer:

Temperature of the gas when flasks are in boiling water = 100 + 273 = 373 K and pressure = 0.5 atm

The average temperature of the gas when one flask is in ice and the other in boiling water

⇒ \(\frac{0+100}{2}=50^{\circ} \mathrm{C}=50+273=323 \mathrm{~K}\)

∴ The temperature of the gas when flasks are in boiling water = 100 + 273 = 373 K and pressure = 0.5 atm The average temperature of the gas when one flask is in ice and the other in boiling water

⇒ \(\frac{0+100}{2}=50^{\circ} \mathrm{C}=50+273=323 \mathrm{~K}\)

Question 22. Equal mass of two gases A and B are kept in two separate containers under the same conditions of temperature and pressure. If the ratio of molar masses of A and JB is 2:3, then what will be the ratio of volumes of the two containers?
Answer:

Let the volume of the container holding gas A = VA and that of the container holding gas B = VB. Suppose, the molar masses of A and B are MA and MB, respectively. Thus, for equal mass (say ‘m’) of each gas, the number of moles of

⇒ \(A\left(n_A\right)=\frac{m}{M_A} \text { and that of } B\left(n_B\right)=\frac{m}{M_B}\)

∴ \(\text { For } A: P V_A=\frac{m}{M_A} R T \text { and for } B: P V_B=\frac{m}{M_B} R T\)

∴ \(\frac{V_A}{V_B}=\frac{M_B}{M_A}=\frac{3}{2}\)

Hence, the ratio of volumes ofthe containers = 3:2

Question 23. At constant pressure for a given amount of gas, will the graphs obtained by plotting V vs t°C and V vs TK be different?
Answer:

According to Charles’ law, V = KT – [1] [at constant pressure for a given mass of a gas]

But, T = 273 + t, hence, V = K(273 + t) -[2] Both the equations [1] & [2] express equations for straight lines. Equation no. [1] represents a straight line passing through the origin. Equation no. [2] does not represent a straight line passing through the origin. The plot of V vs t results in a straight line that cuts the V-axis at 0°C. If this straight line is extrapolated backwards, it meets the y-axis at -273°C.

From the, above two graphs it is evident that there are no actual differences between the two graphs, because -273°C and OK express the same temperature.

Question 24. Under similar conditions of temperature and pressure, if the time taken for effusion of the same volume of H₂, N₂ and CO₂ gas through the same porous wall are t₁, t₂ and t₃  respectively, then arrange t₁, t₂ and t₃ in their increasing order.
Answer:

At constant temperature and pressure, if V volume of H₂ gas effuses in time t j, then according to Graham’s law, \(\frac{V}{t_1} \propto \frac{1}{\sqrt{M_{\mathrm{H}_2}}}\)

For N₂ gas and CO₂ gas, the equations are respectively,

⇒ \(\frac{V}{t_2} \propto \frac{1}{\sqrt{M_{\mathrm{N}_2}}}\)

⇒ \(\frac{V}{t_3} \propto \frac{1}{\sqrt{M_{\mathrm{CO}_2}}}\)

Volume of gas effused in each case is the same From equations [1] & [2]; from equations [2] & [3] we get \(\frac{t_2}{t_1}=\sqrt{\frac{M_{\mathrm{N}_2}}{M_{\mathrm{H}_2}}}; \frac{t_3}{t_2}=\sqrt{\frac{M_{\mathrm{CO}_2}}{M_{\mathrm{N}_2}}}\)

Since, MH₂ < MN₂ < Mc₂, therefore, t₂ > and t₂ > t₂ Hence, t₁,t₂ and t₂ will be in order: t₁<t₂<t₂

Question 25. If, at a given temperature, the total kinetic energy of the molecules in a unit volume of an ideal gas is E, show that the pressure of the gas, P = 2/3E.
Answer:

According to the kinetic gas equation, \(\mathrm{PV}=\frac{1}{3} \mathrm{mnc}^2\)

Suppose, m = mass of each molecule of the gas and n = number of molecules present in V volume of the gas, c = root mean square velocity of gas molecule

∴ \(P V=\frac{2}{3} \times n \times \frac{1}{2} m c^2 \quad \text { or, } P=\frac{2}{3} \times\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2\)

where \(\frac{n}{V}\) the number of molecules per unit volume and \(\frac{1}{2} m c^2\) = the average kinetic energy of each molecule

∴ \(\left(\frac{n}{V}\right) \times \frac{1}{2} m c^2\) hre total kinetic energy of the molecules present per unit volume = E

∴ P = 2/3E (Proved)

Question 26. For the molecules of a given gas at a constant temperature, arrange the most probable velocity (cm), root mean square velocity (CRM) and average velocity (CFL) in the order of their increasing values. With the increase in temperature, will the ratio of these velocities increase, decrease or remain constant? What will the effect of increasing temperature be on the value of (crnls- cm) for a given gas?
Answer:

1. At temperature T, the most probable velocity (cm) ofthe gas molecules = \(\sqrt{\frac{2 R T}{M}}\) [M – molar mass ofthe gas] rms velocity \(\left(c_{r m s}\right)=\sqrt{\frac{3 R T}{M}}\) average velocity \(\left(c_a\right)=\sqrt{\frac{8 R T}{\pi M}}\)

∴ \(c_{r m s}>c_a>c_m\)

2.  Since the value of each of the velocities, crms’ ca and cm’ proportional to Jf, their ratio remains unaltered with temperature rise.

3. Suppose, at temperature, riK, the root mean square velocity (arms) and most probable velocity (cm) of the molecules of a gas are (arms)j and (cm) respectively. If the difference between the values of these velocities is Axx, then

⇒ \(\Delta x_1=\sqrt{\frac{3 R T_1}{M}}-\sqrt{\frac{2 R T_1}{M}}=\sqrt{\frac{R T_1}{M}}(\sqrt{3}-\sqrt{2})\)

Let the temperature of the gas be raised to T2K and at this temperature, the difference between the values of these two velocities is Ax2, then,

⇒ \(\Delta x_2=\left(c_{r m s}\right)_2-\left(c_m\right)_2=\sqrt{\frac{R T_2}{M}}(\sqrt{3}-\sqrt{2})\)

Since T₂ > , Ax₂ > Ax₁. Hence, the difference between the values of these velocities increases with the temperature rise.

Question 27. A and B are closed flasks having the same volume. In flask A, O₂ gas is present at TK and 1 atm pressure. In flask B, H₂ gas is present at 1 atm pressure. If these gases behave ideally, then compare their O total kinetic energies, total number of molecules, and root mean square velocities.
Answer:

Suppose, M₁ and n₂ are the number ofmoles ofthe gases present in flask A and flask B, respectively. Applying the ideal gas equation to the gases O₂ and H₂, we obtain

⇒ \(P V=n_1 R T \text { and } P V=n_2 R \frac{T}{2}\)

∴ \(\frac{n_1}{n_2}=\frac{1}{2}\)

The total kinetic energy of the molecules of mol O₂ in flask A, \(E_1=n_1 \times \frac{3}{2} R T\) and the total kinetic energy of the molecules of n₂ mol H₂ in flask B, \(E_2=n_2 \times \frac{3}{2} R \times \frac{T}{2}\)

[ since Total kinetic energy of the molecules of1mol gas =[Rx absolute temperature]

∴ E₁=E₂

So, the total kinetic energy of the molecules of O₂ gas = the total kinetic energy of the molecules of H₂ gas

If n₁ and n₂ be the number of molecules of O₂ gas and H₂ gas respectively then n1 = nxN and n₂ = n₂xN [AT= Avogadro’s number]

∴ \(\frac{n_1^{\prime}}{n_2^{\prime}}=\frac{n_1 \times N}{n_2 \times N}=\frac{1}{2}\)

∴ Number of molecules in H₂ gas = 2 x number of molecules in oxygen gas

ms velocity of O₂, Crms \(=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R T}{32}}\)

rms velocity of H₂, Cms \(=\sqrt{\frac{3 R \frac{T}{2}}{2}}=\sqrt{\frac{3 R T}{4}}\)

∴ \(\frac{c_{r m s}\left(\mathrm{O}_2\right)}{c_{r m s}\left(\mathrm{H}_2\right)}=\sqrt{\frac{4}{32}}=\sqrt{\frac{1}{8}}=\frac{1}{2 \sqrt{2}}\)

Question 28. Which one of the gases, under the given conditions, exhibits real gas behaviour?

1. 0.25 mol CO₂, T = 1200K, 1. = 24.63 atm, V=1 L
2. 1.0 mol SO₂, T = 300 K, P = 50 atm, V = 0.35 L

Answer:

At a certain temperature and pressure, if the compressibility factor (Z) of a gas is less than or more than 1, then at that temperature and pressure the gas behaves as a real gas. If Z = 1, then the gas exhibits ideal behaviour

1. For CO₂ gas \(Z=\frac{P V}{n R T}=\frac{24.63 \times 1}{0.25 \times 0.0821 \times 1200}=1\)

2. For SO₂ gas ,Z \(=\frac{P V}{n R T}=\frac{50 \times 0.35}{1 \times 0.0821 \times 300}=0.71\)

Under the given conditions, for CO₂ gas, Z = 1 and hence it shows ideal behaviour. In the case of SO₂, Z < 1. So, under the given conditions, SO₂ behaves as a real gas.

Question 29. Write the van der Waals equation for a real gas containing n molecules.
Answer:

If v is the volume of mol of a real gas at a pressure P and temperature T then the van der Waals equation for the gas is

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

Suppose, the volume for a real gas containing n molecules = V. So, volume for n/N moles of real gas = V [N= Avogadro’s number]

Therefore, the volume for 1 mol of real gas \(=\frac{V \times N}{n}=v\)

Substituting the value of v in equation [1], we obtain \(\left(P+\frac{n^2 a}{V^2 N^2}\right)(V N-n b)=n R T\)

This is the van der Waals equation for a real gas containing n molecules

Question 30. What will the value of compressibility factor (Z) be for a gas if the pressure correction term in the van der Waals equation for the gas is neglected?
Answer:

For 1 mol of a real gas \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\)

If the pressure correction term is neglected, then \(P+\frac{a}{V^2} \approx P\)

∴ P(V-b) = RT or, PV -Pb = RT

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

∴ \(Z=1+\frac{P b}{R T}\)

Question 31. The value of van der Waals constant ‘a’ for nitrogen gas is 1.37 L²-atm mol^-2, but that for ammonia gas is 4.30 L_²-atm-mol^-2 What is the reason for this large difference? Which one of these two gases would you expect to have a higher critical temperature?
Answer:

Since N₂ is a non-polar molecule, the only attractive forces that operate between the molecules in N₂ gas are weak London forces. On the otherhand, NH₃ is apolarmolecule. So, the molecules in NH₃ gas experience dipole-dipole attractive forces in addition to London forces.

Hence, the intermolecular forces of attraction in NH₃ gas are much stronger than those in N₂ gas, and this makes the value of ‘a’ for NH₃ gas higher. The critical temperature of a gas depends upon the strength of intermolecular forces of attraction in the gas.

The stronger the intermolecular forces of attraction in a gas, the higher the critical temperature of the gas. Since the intermolecular forces of attraction are stronger in NH₃ gas than those in H₂ gas, the critical temperature of NH₃ will be higher than that of H₂.

Question 32. The values of ‘a’ and ‘b’ for three A real gases B A, Band C Care—

1. Which one of these gases has the largest molecular size?
2. Which one of these will behave most like an ideal gas at STP?

Answer:

1. The volume of a molecule depends on its radius. Now we know b ∝ r₃. Hence, the molecular size will be the largest for the gas which has the highest value of ‘b’. From die given values, it is found that gas B has the highest value of ‘b’ So, so the molecule ofthe gas will be the largest.

2. A gas with small values of ‘a’ and ‘b‘ behaves close to an ideal gas. For gas A, these two quantities have the smallest values. Hence at STP gas A will show the most ideal behaviour.

Question 33. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer:

The equation expressing the relation between density (d), pressure (P), absolute temperature (T) and molar mass (M) of a gas is d \(=\frac{P M}{R T}\)

Density ofthe oxide of the gas, \(d_1=\frac{2 \times M}{R \times 273}\)

Density of N₂ gas, d₂ \(=\frac{5 \times 28}{R \times 273}\)

It is mentioned that under the given conditions, d₁ = d₂

∴ \(\frac{2 \times M}{n \times 273}=\frac{5 \times 28}{R \times 273} \quad \text { or, } M=70\)

Therefore, the molecular mass ofthe oxide is 70 g-mol^-1

Question 34. The pressure of Ig of an Ideal gas A at 27°C Is found to be 2 bar. When 2 g of another ideal gas B Is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses
Answer:

Suppose, the molar masses of gases A and B are M_A and M_B g-mol-1 respectively.

⇒ \(\lg \text { of } A=\frac{1}{M_A} \text { mol of } A \text { and } 2 \mathrm{~g} \text { of } B=\frac{2}{M_B} \text { mol of } B\)

Using the ideal gas equation (PV = NRT) for the gas A and the gas mixture of A and B, we have—

⇒ \(2 \times V=\frac{1}{M_A} R T\)……………………………..(1)

⇒ \(B=\frac{2}{M_B}\)……………………………..(2)

Dividing Equation (2) by (1), we have

1.5 = \(\frac{M_B+2 M_A}{M_B}\)

Or,\(2 M_A=0.5 M_B\)

Or,\(\frac{M_A}{M_B}=\frac{1}{4}\)

Question 35. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and lbar will be released when 0.15g aluminium reacts?
Answer:

The following reaction involving aluminium (Al) and caustic soda (NaOH) produces H₂ gas

⇒ \(\begin{array}{ll}
2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{NaAlO}_2(a q)+ & 3 \mathrm{H}_2(g) \\
2 \times 27 \mathrm{~g} & 3 \times 1 \mathrm{~mol}
\end{array}\)

According to this reaction 54g of Al = 3 mol of H₂

∴ \(0.15 \mathrm{~g} \text { of } \mathrm{Al} \equiv \frac{3}{54} \times 0.15 \equiv 8.33 \times 10^{-3} \mathrm{~mol} \mathrm{H}_2\)

To calculate the volume of the liberated H₂ gas, we apply the ideal gas equation, PV = nRT.

Given:

P = bar and T = (273 + 20)K =293K

The number of moles of liberated H₂ gas (n) =8.33 × 10^-3 mol

∴ \(V=\frac{n R T}{P}=\frac{8.33 \times 10^{-3} \times 0.0821 \times 293}{0.987} \mathrm{~L}=0.203 \mathrm{~L}=203 \mathrm{~mL}\)

Question 36. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9dm3 flask at 27°C?
Answer:

Total number of moles (n) in the mixture

⇒ \(\left(\frac{3.2}{16}+\frac{4.4}{44}\right)\) 0.3 mol

[Molar mass: \(\mathrm{CH}_4 \Rightarrow 16, \mathrm{CO}_2 \Rightarrow 44\)

As given, V = 9 dm3, T = (273 + 27)K = 300K, P = ?

So, \(p=\frac{n R T}{V}=\frac{0.3 \times 0.0821 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}{9 \mathrm{dm}^3}\)

= 0.821 aztm [ 1=1dm³]

= 8.316 × 10⁴

pa[1 atm=1.013 × 10⁵pa]

Question 37. What will be the pressure of the gaseous mixture when 0.5L of H₂ at 0.8 bar and 2.0L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
Answer:

To calculate the number of moles of unmixed H₂ and O₂ m gases, we apply the ideal gas equation, PV = nRT

In the case of H₂, P = 0.8bar, V = 0.5L and T = (273 + 27)K = 300K

And in case of O₂, P = 0.7bar, V = 2.0L and T = (273 + 27)K = 300K

∴ \(n_{\mathrm{H}_2}=\frac{P V}{R T}=\frac{0.8 \times 0.5}{R \times 300}\)

= \(\frac{0.4}{300 R} \mathrm{~L} \cdot \text { bar }\) and

⇒ \(n_{\mathrm{O}_2}=\frac{P V}{R T}=\frac{2 \times 0.7}{R \times 300}\)

=\(\frac{1.4}{300 R} \mathrm{~L} \cdot \text { bar }\)

In the mixture of H₂ and O₂, the total number of mol

= \(n_{\mathrm{H}_2}+n_{\mathrm{O}_2}\)

=\(\frac{1}{300 R}(0.4+1.4) \mathrm{L} \text { bar }\)

=\( \frac{1.8}{300 R} \mathrm{~L} \cdot \text { bar }\)

For this mixture V = 1L and T = (273 + 27)K = 300K

If the pressure of the mixture is P, then

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

= \(\frac{1.8}{300 R} \times \frac{R \times 300}{1}\) bar

=1.8bar

Question 38. The density of a gas is found to be 5.46 g/dm₃ at 27°C and 2 bar pressure. What will be its density at STP?
Answer:

Density (d), pressure (P), absolute temperature (T) and molar mass of a gas (M) are related by,

d = PM/RT

Under the conditions of T = (273 + 27)K = 300K and

P = 2 bar, the density of the gas

(d) = 5.46 g.dm^-3

∴ = 5.46 g.dm^-3 = \(\frac{(2 \times 1.013) \mathrm{atm} \times M}{0.0821 \mathrm{~atm} \cdot \mathrm{dm}^3 \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}\)

M = 66.37 g.mol¯¹

At STP, if the density of the gas is d₁, then

⇒ \(d_1=\frac{P M}{R T}=\frac{1 \times 66.37}{0.0821 \times 273}\)

At STP, T = 273K P =  1 atm

= \(2.96 \mathrm{~g} \cdot \mathrm{L}^{-1}\)

= \(2.96 \mathrm{~g} \cdot \mathrm{dm}^{-3}\)

Question 39. 4.05mL of phosphorus vapour weighs 0.0625g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer:

Suppose, the molar mass of phosphorus =M g-mol^-1

So, 0.0625g of phosphorus

= \(\frac{0.0625}{\mathrm{M}} \mathrm{mol}\) of phosphorus As given, P = 0.1 bar, T = (273 + 546)K

= 819K, V = 34.05 mL = 34.05 × 10^-3L

Therefore, 0.1 bar × 34.05 × 10^-3L.

⇒\(\frac{0.0625}{M} \mathrm{~mol} \times 0.082 \mathrm{~L} \cdot \mathrm{bar} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 819 \mathrm{~K}\)

or, M = 1232.71, Hence, Molar mass = 1232.71g.mol^-1

Question 40. A student forgot to add the reaction mixture to the round-bottomed flask at 27°C but instead, he/she placed the flask on the flame. After a lapse of time, he realised his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer:

Suppose the number of mol of air inside the flask at 27°C is n1 and that at 477°C is n₂. Since the flask is opened to air, the pressure of air inside the flask at 27°C and 477°C is the same as that of the atmospheric pressure. Let this pressure be P. Again on heating the volume of a round bottom flask remains the same. Hence, the volume ofthe flasks the some at both 27°C and 477°C. Let this volume be V.

Applying the ideal gas equation we have, PV= n₁ R(273 + 27)

= HJ × 300R n₂PV = n₂ (273 + 477) = n₂ × 750R

Hence, n₂ × 300R = n₂ × 750R

Or, n₂ = \(\frac{2}{5} n_1\)

∴ Fraction of air that would have been expelled out = \(\frac{n_1-\frac{2}{5} n_1}{n_1}=\frac{3}{5}\)

Question 41. Payload Is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kgs filled with helium at 1.66 bar at 27°C. (R = 0.083 bar. dm³. K mol^-1 and density of air = 1.2kg.m^-3
Answer:

Volume (V) of the balloon =\(\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times(10)^3 \mathrm{~m}^3\)

= 4186.66m³

= 4186.66 × 10³ dm³

To calculate the number of moles of He gas enclosed in the balloon, we apply the ideal gas equation, PV = nRT

Given: P = 1.66 bar, T = (273 + 27)K = 300K So, according to the equation, PV = nRT,

n = \(\frac{P V}{R T}=\frac{1.66 \times 4186.66 \times 10^3 \mathrm{bar} \cdot \mathrm{dm}^3}{0.083 \mathrm{bar} \cdot \mathrm{dm}^3 \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 300 \mathrm{~K}}\)

= \(279.11 \times 10^3 \mathrm{~mol}\)

Hence, the mass of 1<< gas enclosed in the balloon

= 4  × 279. 11 × 10³g = 1,110 × 10⁶g 1116kg

Therefore, the mass of the balloon filled in with He gas

= (100+ 1116)1kg = 1216kg

The volume of air displaced by the balloon = Volume of the balloon

= 4186.66 × 10³ dm³

Density of air – 1,2 kg. m^-3 = 1.2 × 10^-3 kg.dm^-3

The mass of air displaced by the balloon

= 4180,60 × 10³  × 1,2 × 10^-3 kg =5024kg

Hence, payload = the mass of displaced air- the mass of the balloon = (5024- 1216)kg = 3808 kg

Question 42. Calculate the volume occupied by 8.8 g of C02 at 31.1 °C &1 bar pressure. t=0.083 bar -L-K^-1mol^-3
Answer:

⇒ \(8.8 \mathrm{~g} \mathrm{CO}_2=\frac{8.8}{44}=0.2 \mathrm{~mol} \mathrm{CO}_2\)

Applying the ideal gas equation (PV = NRT) to calculate volume, we have

⇒  \(V=\frac{n R T}{P}=\frac{0.2 \times 0.083 \times(273+31.1)}{1}=5.048 \mathrm{~L}\)

∴ The volume of 8.8 g CO₂ at 31.1°C and 1 bar pressure = 5.048L.

Question 43. 2.9 g of gas at 95°C occupied the same volume as 0.184g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Answer:

Let the molar mass ofthe unknown gas =M g-mol^-1 Number ofmoles for 2.9g ofthe gas \(=\frac{2.9}{M} \mathrm{~mol}\)

Number of moles for 0.184g of H₂ gas \(=\frac{0.184}{2}=0.092 \mathrm{~mol}\)

As both the gases have the same volume at the same pressure and specified temperature, they will have the same value of PV. To calculate PV, we apply the equation

⇒  PV = nRT. For unknown gas \(P V=\frac{2.9}{M} R(273+95)\)

And for He Gas, Pv [= 0.0932R(273+17)

Therefore \(\frac{2.9 \mathrm{R}}{M} \times 368=0.092 R \times 290 \quad \)

Hence, the molar mass ofthe unknown gas = 40g-mol^-1

Question 44. Oxygen gas is present in a 1L flask at a pressure of 7.6 x 10¯1° mm Hg at 0°C. Calculate the number of O₂ molecules in the flask
Answer:

P= \(\frac{7.6 \times 10^{-10}}{760}\)

∴ \(n=\frac{P V}{R T}=\frac{7.6 \times 10^{-10} \times 1}{760 \times 273 \times 0.082}\)

[V = 1L, T = 273K, n = 4.46× 10^-14]

Question 45. Sketch PIT versus T plot for an ideal gas at constant volume. Indicate the value ofthe slope (mass fixed). Under the same conditions of temperature and pressure NH₃, Cl₂ and CO₂ gases are allowed to diffuse through a porous wall. Arrange these gases in the increasing order ofthe rate of diffusion.
Answer:

According to Gay-Lussac’s law, \(\frac{P}{T}=K\) (constant), when the mass and volume of a gas are constant. Hence, the plot of \(\frac{P}{T}=K\) versus T indicates a straight line parallel to the X -axis and the slope ofthe curve is tan 0 = 0.

The order of molar masses of NH₃ Cl₂ and CO₂ gases are—NH₃ < CO₂ < Cl₂.

Hence, the increasing order of their rates of diffusion is Cl₂ < CO₂ < NH₂.

The value of ‘R ‘ in J-K^-1-mol^-1 unit is 8.314

Question 46. A gas of molar mass 84.5g/mol is enclosed in a flask at 27°C has a pressure of 2atm. Calculate the density of the gas. [R = 0.082L.atm- K^-1.mol^-1 ]
Answer:

The density of a gas,

d =  \(\frac{P M}{R T}=\frac{2 \times 84.5}{0.082 \times 300} \mathrm{~g} \cdot \mathrm{L}^{-1}\)

=\(6.87 \mathrm{~g} \cdot \mathrm{L}^{-1}\)

Question 47. For which property of the liquid the shape of a liquid drop is spherical? A 10-litre volumetric flask contains 1 gHe and 6.4 g of O₂ at 27°C temperature. If the total pressure of the mixture is 1.107 atm, then what is the partial pressure ofHe and O₂?
Answer:

Surface tension.

⇒ \(n_{\mathrm{He}}=\frac{1}{4}=0.25 \mathrm{~mol}, n_{\mathrm{O}_2}=\frac{6.4}{32}=0.2 \mathrm{~mol}\)

∴ Total number of moles =(0.25 + 0.2) = 0.45

⇒\(x_{\mathrm{He}}=\frac{0.25}{0.45}=0.5555 \text { and } x_{\mathrm{O}_2}=\frac{0.2}{0.45}=0.4444\)

pHe = 0.5555 × 1.107 atm = 0.615 atm

Po2 = 0.4444 × 1.107 atm =0.491 atm

Question 48. What will be the ratio of \({ }^{235} \mathrm{UF}_6\) diffusion of And \({ }^{235} \mathrm{UF}_6\)
Answer:

Let, the rate of diffusion of \({ }^{235} \mathrm{UF}_6 \text { and }{ }^{238} \mathrm{UF}_6\) be r1 and r2.

According to Graham’s law, \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)

⇒ \(M_1=\text { molecular mass of }{ }^{235} \mathrm{UF}_6=349 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

⇒ \(M_2=\text { molecular mass of }{ }^{238} \mathrm{UF}_6=352 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

∴ \(\frac{r_1}{r_2}=\sqrt{\frac{352}{349}}=1.004\)

Question 49. Determine the volume of 2.2 g of carbon dioxide at 27°C and 570 mmHg pressure.
Answer:

⇒ \(2.2 \mathrm{~g} \text { of } \mathrm{CO}_2 \equiv \frac{2.2}{44}=0.05 \mathrm{~mol} \text {. }\)

According to the ideal gas equation (taking CO2 as an ideal gas), PV = nRT

⇒ \(\text { or, } \frac{570}{760} \times V=0.05 \times 0.082 \times 300\)

or, v= 1.64 The volume of CO₂ at the given condition is 1.64L.