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The First Law Of Thermodynamics

The first law of thermodynamics is an extension of the law of conservation of energy. The law of conservation of energy states that energy can neither be created nor destroyed. It can only be converted from one form to another in an equivalent amount.

The First Law Of Thermodynamics Can Be Stated In Different Ways: The total energy of the universe is always constant. or, Energy can neither be created nor be destroyed. It can only be transformed from one form to another in an equivalent quantity.

Whenever a definite quantity of one form of energy disappears, an exactly equivalent amount of another form of energy appears. or, The energy of an isolated system is always constant.

Justification Of The First Law Of Thermodynamics

There is no theoretical proof of the first law of thermodynamics. Human experience of the macroscopic behavior of a large number of systems is indeed the basis of the law. The law, when applied to various fields, is valid.

NCERT Class 11 Chemistry Notes on First Law of Thermodynamics

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Equivalence Of Work And Heat: From different experiments it has been observed that heat and work are always equivalent. Irrespective of the method by which work is performed, a definite quantity of heat is obtained in exchange for a definite quantity of work. Scientist Joule proved that 1 calorie of heat is produced in exchange for 4.1843 J of work.

So 4.184J of work is equivalent to 1 cal of heat. Therefore, creation or destruction of energy is not possible. When one form of energy disappears exactly an equivalent amount of another form ofenergy is obtained.

Perpetual Motion Machine: The impossibility of constructing a perpetual motion machine of the first kind supports the first law A perpetual motion Is a type that can work In a cyclic process and produce work without consuming any energy or producing more work by consuming lost energy.

All efforts In making such machines are In vain. Hence, it can be concluded that energy can neither be created nor be destroyed, but one form of energy can be converted Into an equivalent quantity of another form of energy.

First Law of Thermodynamics Class 11 Chemistry Notes

The Total Energy Of A System Is Constant: Since an Isolated system does not Interact with Its surroundings, it cannot exchange either matter or energy with Its surroundings. This is why the energy of the system always remains constant.

The Mathematical Form Of The First Law Of Thermodynamics

A closed system changes its state if it exchanges heat or work or both with its surroundings. Consequently, the internal energy of the system also changes. Suppose, a closed system has an internal energy of Ux. Let q amount of heat be added to the system.

This increases the internal energy of the system to a value of ( Ux + q). Now, if an amount of work w is done on the system, then the internal energy of the system will further increase and will become (Ul+q + w). Suppose, the internal energy of the system at the final state is U₂. So, U₂ = Ux + q + w or, U₂– t₁ = q + w

∴ AU=q+w

[AU = U2-U1 = the change in internal energy of the system] vi i.e., the change in internal energy of a closed system = heat absorbed by the system + work done on the system. The equation is the mathematical formulation of the first law of thermodynamics. Important points regarding the equation, AU = q + wi]

Class 11 Chemistry First Law of Thermodynamics Notes

Unit of AU, q, or w: In this equation, AU, q, and w must be expressed in the same unit such as joule or calorie.

IUPAC sign convention of q and w: w = positive: work is done on the system. w = negative: work is done by the system. q = positive: heat is absorbed by the system. q = negative: heat is lost by the system

Illustration: In a process, If a system absorbs 140 1 of heat and performs 80 J of work, then q = +140 J and w = -80 I.

In this case the change in internal energy, AU = q + w – 140- 80 = 60 J.0 In a process, if a system rejects 2001 heat and performs 60 J of work, then q = – 200 I and w = – 60 J. In this case, the change in internal energy, AU = q+w = -200-60 = -260 J.

Although q and w are not state functions, the sum of these two quantities does not depend upon the path. This Is because the sum of q and w is equal to AU, which is path-independent.

The mathematical expression of the first law of thermodynamics for an infinitesimal change: According to the first law of thermodynamics, for a closed system, AU = q + w; where q = heat absorbed by the system, w = work done on the system, A U = change in internal energy.

CBSE Class 11 First Law of Thermodynamics Notes

In Case Of An Infinitesimal Change, The Above Equation Can Be Written As: dU = 6q + 5w where 5q = infinitesimal amount of heat absorbed by the system, 8w = infinitesimal amount of work done on the system, dU = infinitesimal change in the internal energy. Equation [1] represents the mathematical form of the first law of thermodynamics for infinitesimal changes.

The Mathematical Form Of The First Law Of Thermodynamics In Case Of A Process That Involves Only P-Work: According to the first law, AU = q+w. In this equation, w represents the sum of all types of work (mechanical, electrical, magnetic work, etc.). If only P-Vwork is performed, then \(w=-\int_{V_1}^{V_2} P_{e x} d V \quad\left[P_{e x}=\text { external pressure on the system }\right]\)

V1 and V2 are the initial and final volumes of the system respectively. When only P-V work is performed, the mathematical form ofthe first law of thermodynamics is

⇒ \(\Delta U=q-\int_{V_1}^{V_2} P_{e x} d V\)

If the volume of the system increases, then V₂ > V₁ (expansion). In this case, work is done by the system of the surroundings against the external pressure (Pex).

If the volume of the system is decreased by the external pressure (Pa) then V2<VX (compression). In this case, work is done on the system by the surroundings.

If the die volume of the system increases from V₁ to V₂ against a constant external pressure (PfX). then AU = q-Pex{Vs-V{)=q-PexV <Va>

If the volume of the system is decreased From V₁ to V₂ bv a constant external pressure (Ppv), then A U = q-P1V₂-V₂)=q-PexAV(V₂< V₂) isolated system and the first law of thermodynamics: According to the first law, AU = q + w. An isolated system does not interact with its surroundings. So such a system does not exchange heat (q) or work (w) with its surroundings and hence, q = 0 and w = 0. Therefore, the internal energy of an isolated system remains constant (AC/ = 0)

First Law of Thermodynamics NCERT Notes Class 11

Class 11 Chemistry Chemical Thermodynamics  Numerical Examples

Question 1. The volume and temperature of 2mol of an ideal gas 10L & arc, respectively. The gas is allowed to expand in an isothermal reversible process to attain a final volume of 25 L. Calculate the maximum work done
Answer: We know that maximum work is obtained in a reversible isothermal expansion
Now, work done by an ideal gas in an isothermal reversible expansion, \(w=-n R T \ln \frac{V_2}{V_1}\) Given: n = 2, T = (273 + 27)=300K. V1= 10 L, V2=25L

∴ \(w=\left[-2 \times 8.314 \times 300 \ln \frac{25}{10}\right] \mathrm{J}=-4570.82 \mathrm{~J}\)

∴ The maximum work done bythe gas = 4570.82 J

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Question 2. A 3 mol sample of an ideal gas at STP expands in an isothermal reversible process to attain a final volume of 100L. Calculate the work done by the gas.
Answer: The volume of1 mol of an ideal gas at STP – 22.4 L. So, at STP the volume of 3 mol of an ideal gas =3×22.4 = 67.2 L

Now, work done by an ideal gas in an isothermal reversible expansion process, \(w=-n R T \ln \frac{V_2}{V_1}\)

Given: n = 3, T= (273 + 0) =273K, V1 = 67.2L, V2 =100L

∴ \(w=-3 \times 8.314 \times 273 \ln \frac{100}{\bar{\sigma} .2}=-2706.62 \mathrm{~J}\)

∴ Work done bythe gas = 2706.62 J.

Question 3. A 23 mol sample of an ideal gas is compressed in i reversible isothermal process from a volume of 12 to a volume of 5 L at 27°C. Calculate the work done on the gas.
Answer: In a reversible isothermal compression of an ideal gas, the work done on the gas, w \(=-n R T \ln \frac{V_2}{V_1}\left[\text { where } V_2<V_1\right]\)

Given: n = 2.5, T= (273 + 27) = 300K, V₁=121., V2=5L

∴ \(w=-2.5 \times 8.314 \times 300 \ln \frac{5}{12}=5458.98 \mathrm{~J}\)

∴ Work done on the gas = 5458.98J.

Class 11 Chemistry Exemplar Solutions Chemical Thermodynamics

Question 4. A gas is compressed by an external pressure of atm. The work done in the process is 1034 J. How much volume of the gas is reduced?
Answer: Workdone on the gas due to compression

⇒ \(w=-P_{e x} \Delta V=-P_{e x}\left(V_2-V_1\right)=P_{e x}\left(V_1-V_2\right)\)

Where Pex = external pressure; V1 and V2 are the initial and final volumes ofthe gas respectively.

As per the given data, \(w=1034 \mathrm{~J}=\frac{1034}{101.3}=10.2073 \mathrm{~L} \cdot \mathrm{atm}\)

and Pex = 5 atm ∴ 1 L-atm = 101.3 J

∴ 10.2073 = 5 X ( V1- V2)

∴ V₁-V₂ = 2.04 L

∴ The decrease in volume of the gas = 2.04l.

Question 5. Work done by 3 mol of an ideal gas in an isothermal reversible expansion at 30°C is 9.5 kJ. If the initial volume of the gas is 20 L, then what will be the final volume of the gas?
Answer: Work done by an ideal gas in an isothermal reversible expansion \(w=-n R T \ln \frac{V_2}{V_1}\) As per given data, n vi = (273 + 30)K = 303K and w = -9.5 kj = -9500 J. Initial volume ofthe gas, V1 = 20 L; Finalvoiume, V2=?

∴ \(-9500=-3 \times 8.314 \times 303 \ln \frac{V_2}{20} \text { or, } V_2=70.3 \mathrm{~L}\)

∴ After expansion, the final volume ofthe gas = 703 L.

Question 6. The pressure of 3 mol of an ideal gas is 10 atm at 27°C. Calculate work done by the gas when it is expanded isothermally against an external pressure of1 atm
Answer: As per the given data, the initial pressure of the gas, P₁= 10 atm, and the gas is expanded against an external pressure of 1 atm. Therefore, the final pressure, P₂ = 1 atm. Since the external pressure is very much Jess than the initial pressure, the gas will expand irreversibly and after expansion, the pressure of the gas will be equal to the external pressure i.e., 1 atm. In an isothermal irreversible expansion

⇒ \(w=-n R T\left(1-\frac{P_2}{P_1}\right)\)

Here, n = 3, T = (273 + 27)K = 300K

∴ \(w=-3 \times 8.314 \times 300\left(1-\frac{1}{10}\right)=-6734.34 \mathrm{~J}\)

Therefore, work done by the gas in this isothermal irreversible expansion = 6734.34 J.

Question 7. The pressure of 6 mol N₂ gas kept in a cylinder is 30 atm at 30°C. Suddenly the gas comes out of the cylinder due to leakage. If the atmospheric pressure and temperature are 1 atm and 30°C, then calculate the work done by the gas. Assume that the gas behaves ideally.
Answer: The atmospheric pressure (1 atm) is very much less than the initial pressure of the gas in the cylinder (30 atm) and the temperature of the gas is the same as the atmospheric temperature. So, the gas will expand isothermally and irreversibly against an external pressure of1 atm.

In an isothermal irreversible expansion, \(w=-n R T\left(1-\frac{P_2}{P_1}\right)\)

As per given data, P1 = 30 atm, P2 = 1 atm, n = 6 and T = (273 + 30) = 303K

∴ \(w=-6 \times 8.314 \times 303\left(1-\frac{1}{30}\right)=-14611.02 \mathrm{~J} .\)

Chemical Thermodynamics Class 11 Exemplar Solutions

Question 8. A 5 mol sample of an ideal gas is compressed isothermally and irreversibly from 1.5 atm to 15 atm at 27 °C. Calculate the work done on the gas in the calorie unit.
Answer: Work done in an isothermal irreversible expansion, \(w=-n R T\left(1-\frac{P_2}{P_1}\right)\)

As per given data, n = 5 , Px = ,1.5atm, P2 = 15 atm, T = (273 + 27)K = 300K

∴ \(w=-5 \times 1.987 \times 300 \times\left(1-\frac{15}{1.5}\right)=26824.5 \mathrm{cal}\)

∴ The work done on the gas = 26.82 kcal.

Question 9. What amount of work is done if an ideal gas expands from 10L to 20L at 2 atm pressure?
Answer: We know, w = -Pex(V₂-v₁).

Given: Pex = 2 atm, = 10 L and V₂ = 20 L

∴ w = -2(20-10) L-atm= -20 L-atm

=-20 x 101.3J = -2026J

since 1 L-atm = 1013 J

A negative value of w indicates work is done by the gas (system). Therefore, the magnitude of work done by the gas (system) = 2026 J.

Question 10. What amount of work is done if an ideal gas is compressed from 0.5 to 0.25L under 0.1 atm pressure?
Answer: We know, w = —Pex(.V₂ — Y{) [In case of compression V₂<V₁]

Given: Pgx = 0.1 atm, V₁ = 0.5 L and V2 = 0.25

∴ w =-0.1(0.25- 0.5) = 0.025 L-atm

= 0.025 X 101.3J = 2.53 J

Therefore, work done on the gas = 2.53 J

Question 11. Calculate the work done when 1 mol of water vaporizes at 100°C and atm pressure. Assume water vapor behaves like an ideal gas
Answer: 1 mol water,100C , 1 atm – 1 mol water vapour 100C, 1 atm

Pex = external pressure = 1 atm, V₁ and V₂ are volumes of1 mol of water and 1 mol of water vapor, respectively. The volume of1 mol of water vapor is very much greater than that of1 mol water. Thus, (V2-V1) and it makes(V₂– V₁) = V₂.

∴ Work done, w = -PexV₂

If water vapour behaves like an ideal gas, then PexV₂ = PT amount of water vapour = 1 mol

∴ w = -PexV₂ = -RT

Here, T = (273 + 100)K = 373K

W = -RT = -8.314 X 373 = -3101.12

Therefore, the amount of work done by the system = 3101.12 J.

CBSE Exemplar Class 11 Chemistry Chemical Thermodynamics Solutions

Question 12. ron reacts with dilute HCI quantitatively to form \(\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_2(a q)+\mathrm{H}_2(g)\) 56g of iron is allowed to react completely with dil HCI at 25 °C. If this reaction is carried out separately in a closed container of fixed volume and an open breaker, then calculate the work done in each case. Assume If2 gas behaves like an ideal gas.
Answer: Work done, w \(-P_{e x} \Delta \mathrm{V}=-P_{e x}\left(V_2-V_1\right)\)

Pex – Constant external pressure, V₁ and V₂ are the initial and final volumes, respectively.

When the reaction occurs in a closed container of fixed volume, the change in volume of the system, AV = 0, hence w = 0.

In this case, the volume of the system increases against atmospheric pressure due to the formation of hydrogen gas in the reaction.

⇒ \(\begin{aligned}
& \text { So, } w=-P_{e x} \Delta V=-P_{e x} V_{\mathrm{H}_2}=-n_{\mathrm{H}_2} R T \\
& \text { Reaction: } \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_2(a q)+\mathrm{H}_2(g) \\
& 56 \mathrm{~g}(1 \mathrm{~mol}) \quad 1 \mathrm{~mol}
\end{aligned}\)

∴ nH2 = 1. Given: T = (273 + 25) = 298K

∴ w = -1 X RT = -8.314 x 298 = -2.477 kj

Therefore, work done by the system = 2.477 kj.