Thermodynamics | Questions based on Entropy

Questions

For an ideal gas C_p = 2.5R, calculate the change in entropy when 3 moles of this gas is heated from 300 K to 600 K at: (a) Constant pressure (b) Constant volume.

$$Givenvalues:n=3,C_p=2.5R,T_1=300K,T_2=600K$$

$$C_v=C_p-R=2.5R-R=1.5R$$

(a) Change in entropy at constant pressure:

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$$

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}$$

$$\Delta S=3\times 2.5R\times ln\frac{600}{300}$$

$$\Delta S=7.5\times 8.314\times ln2$$

$$\Delta S=43.22J{K}^{-1}$$

(b) Change in entropy at constant volume:

$$\Delta S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$$

$$\Delta S=n{C}_{v}ln\frac{T_2}{T_1}$$

$$\Delta S=3\times 1.5R\times ln\frac{600}{300}$$

$$\Delta S=4.5\times 8.314\times ln2$$

$$\Delta S=25.93J{K}^{-1}$$

Calculate the change in entropy when 3 moles of an ideal gas is heated and compressed simultaneously from 300 K, 1 atm to 400 K, 5 atm. (C_p = 7R/2)

Applying the formula of ΔS:

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$$

$$\Delta S=3\times \frac{7R}{2}ln\frac{400}{300}+3\times Rln\frac{1}{5}$$

$$\Delta S=3\times 3.5\times 8.314\times ln\frac{4}{3}+3\times 8.314\times ln\frac{1}{5}$$

$$\Delta S=-15.03J{K}^{-1}$$

Since, ΔS is negative, this process is non-spontaneous.

By how much does the entropy of 3 moles of an ideal gas change if pressure is reduced from 2 bar to 1 bar without any change in temperature?

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$$

Since, temperature is constant, we can write:

$$\Delta S=nRln\frac{P_1}{P_2}$$

$$\Delta S=3\times 8.314\times ln\frac{2}{1}$$

$$\Delta S=17.288J{K}^{-1}$$

Evaluate the change in entropy when 3 moles of an ideal gas is heated from 27^oC to 727^oC at constant pressure of 1 atm. (C_p = 23.7 J K^-1mol^-1)

$$Given:T_1=(27+273)=300K$$

$$and,T_2=(727+273)=1000K$$

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$$

Since, pressure is constant, we can write:

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}$$

$$\Delta S=3\times 23.7\times ln\frac{1000}{300}$$

$$\Delta S=85.6J{K}^{-1}$$

Nitrogen is heated from 300 K to 600 K at constant pressure of 1 atm. Calculate the change in molar entropy if C_p = 20 + 10 T, where T = temperature.

$$\Delta S=\int \frac{d{q}_{rev}}{T}$$

$$\Delta S=\underset{T_1}{\overset{T_2}{\int }}\frac{n{C}_{p}dT}{T}$$

Molar change in entropy means change in entropy per mole. So, molar entropy change ΔS_m is given by:

$$\Delta S_m=\frac{\Delta S}{n}$$

$$\Delta S_m=\underset{T_1}{\overset{T_2}{\int }}\frac{C_{p}dT}{T}$$

$$\Delta S_m=\underset{T_1}{\overset{T_2}{\int }}\frac{20+10T}{T}dT$$

$$\Delta S_m=\underset{T_1}{\overset{T_2}{\int }}\frac{20}{T}dT+\underset{T_1}{\overset{T_2}{\int }}\frac{10T}{T}dT$$

$$\Delta S_m=20ln\frac{T_2}{T_1}+10(T_2-T_1)$$

$$\Delta S_m=20ln\frac{600}{300}+10(600-300)$$

$$\Delta S_m=20ln2+10\times 300$$

$$\Delta S_m=3013.863J{K}^{-1}mo{l}^{-1}$$

A five litre container is divided so that 1 L of O₂ gas at 1 atm and 4 L of N₂ gas at 2 atm are on the two sides of the thin membrane. The membrane is then broken, allowing the gas to mix up at 300 K. Calculate the entropy of mixing for this process.

• Moles of O₂ gas is given by:

$$n_{O_2}=\frac{PV}{RT}$$

$$n_{O_2}=\frac{1\times 1}{0.0821\times 300}=0.0406$$

• Moles of N₂ gas is given by:

$$n_{N_2}=\frac{PV}{RT}$$

$$n_{N_2}=\frac{2\times 4}{0.0821\times 300}=0.3248$$

• When two gases are mixed, mole fractions of O₂ and N₂ are:

$${{\rm X}}_{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{N_2}}$$

$${{\rm X}}_{O_2}=\frac{0.0406}{0.0406+0.3248}=0.11$$

$${{\rm X}}_{N_2}=\frac{n_{N_2}}{n_{O_2}+n_{N_2}}=0.89$$

• Entropy change for O₂:

$$\Delta S_{O_2}=n_{O_2}Rln{{\rm X}}_{O_2}$$

$$\Delta S_{O_2}=0.0406\times 8.314\times ln(0.11)=-0.745J{K}^{-1}$$

• Entropy change for N₂:

$$\Delta S_{N_2}=n_{N_2}Rln{{\rm X}}_{N_2}$$

$$\Delta S_{N_2}=0.3248\times 8.314\times ln(0.89)=-0.315J{K}^{-1}$$

• Total entropy change or entropy of mixing is given by:

$$\Delta S=\Delta S_{O_2}+\Delta S_{N_2}$$

$$\Delta S=(-0.745)+(-0.315)=-1.06J{K}^{-1}$$

Derive the condition to obtain maximum molar entropy change if two gases are mixed to form a binary mixture.

• Let the two gases be denoted by A and B. Let the mole fraction of A = Χ_A and mole fraction of B = 1 - Χ_A. Molar entropy change of mixing of these two gases:

$$\Delta S_m={{\rm X}}_{A}ln{{\rm X}}_A+(1-{{\rm X}}_A)ln(1-{{\rm X}}_A)$$

$$\Delta S_m={{\rm X}}_{A}ln{{\rm X}}_A+ln(1-{{\rm X}}_A)-{{\rm X}}_{A}ln(1-{{\rm X}}_A)$$

• Differentiating with respect to χ_A:

$$\frac{d(\Delta S_m)}{d{{\rm X}}_A}=ln\frac{{{\rm X}}_A}{(1-{{\rm X}}_A)}$$

• For maximum value of ΔS_m:

$$\frac{d(\Delta S_m)}{d{{\rm X}}_A}=0$$

$$ln\frac{{{\rm X}}_A}{(1-{{\rm X}}_A)}=0$$

$$\frac{{{\rm X}}_A}{(1-{{\rm X}}_A)}=1$$

$$2{{\rm X}}_A=1$$

$${{\rm X}}_A=0.5$$

• Hence, to obtain maximum molar entropy change by mixing two gases A and B, mole fraction of A, Χ_A = 0.5 and mole fraction of B, Χ_B = 0.5, ie, A and B should be equimolar or they should have same number of moles.

One mole of an ideal gas is expanded isothermally at 298 K until its volume is tripled. Find ΔS_gas and ΔS_total under following conditions: (a) Expansion is reversible (b) Expansion is irreversible where 836.8 J less heat is absorbed than in the first (c) Free expansion

• Let's calculate the entropy change for the system or gas.

$$\Delta S_{gas}=nRln\frac{V_2}{V_1}$$

$$\Delta S_{gas}=nRln\frac{3V}{V}$$

$$\Delta S_{gas}=nRln(3)$$

$$\Delta S_{gas}=1\times 8.314\times ln(3)$$

$$\Delta S_{gas}=9.134J{K}^{-1}$$

This is the value of ΔS_gas for all three cases mentioned in the question.

• (a) Since, the process is reversible, ΔS_total = 0. So, ΔS_surrounding = -ΔS_gas = -9.134 JK^-1

• (b) ΔS_surrounding for irreversible process is given by:

$$\Delta S_{surrounding}=\frac{-q_{irr}}{T}$$

$$\Delta S_{surrounding}=\frac{-(q_{rev}-836.8)}{T}$$

ΔS_total is given by:

$$\Delta S_{total}=\Delta S_{gas}+\Delta S_{surrounding}$$

$$\Delta S_{total}=\frac{q_{rev}}{T}+\frac{-(q_{rev}-836.8)}{T}$$

$$\Delta S_{total}=\frac{836.8}{T}$$

$$\Delta S_{total}=\frac{836.8}{298}=2.808J{K}^{-1}$$

• (c) For free expansion, ΔS_surrounding = 0.

$$So,\Delta S_{total}=\Delta S_{gas}=9.134J{K}^{-1}$$

Calculate the entropy change when 0.5 L of an ideal gas (C_v = 12.6 JK^-1mol^-1) at 300 K and 1 atm pressure is allowed to expand to double its volume and simultaneously heated to 373 K.

Number of moles, n can be calculated by ideal gas equation:

$$n=\frac{PV}{RT}$$

$$n=\frac{1\times 0.5}{0.0821\times 300}=0.02$$

Change in entropy, ΔS is given by:

$$\Delta S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$$

$$\Delta S=0.02\times 12.6\times ln\frac{373}{300}+0.02\times 8.314\times ln\frac{2V_1}{V_1}$$

$$\Delta S=0.17J{K}^{-1}$$

The temperature of an ideal monoatomic gas is raised from 25^oC to 250^oC. What must be the accompanying pressure change in order that the entropy of gas may be unaffected by the complete process?

Given values: T₁ = 25 + 273 = 298 K, T₂ = 250 + 273 = 523 K

As per the question, ΔS = 0.

$$n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}=0$$

$$n{C}_{p}ln\frac{T_2}{T_1}=-nRln\frac{P_1}{P_2}$$

$$\frac{5R}{2}ln\frac{T_2}{T_1}=-Rln\frac{P_1}{P_2}$$

$$\frac{5}{2}ln\frac{523}{298}=ln\frac{P_2}{P_1}$$

$$2.5ln(1.755)=ln\frac{P_2}{P_1}$$

$$ln(1.755{)}^{2.5}=ln\frac{P_2}{P_1}$$

$$4=\frac{P_2}{P_1}$$

$$P_2=4P_1$$

So, final pressure is four times of the initial pressure.

By how much does the entropy of 3 moles of an ideal gas change in going from pressure of 2 bar to a pressure of 1 bar without any change in temperature? If the surrounding too is at 1 bar pressure and 300 K temperature and the expansion is against the constant external pressure of surroundings, tell whether the process is reversible or irreversible.

ΔS_system is given by:

$$\Delta S_{system}=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$$

$$\Delta S_{system}=nRln\frac{P_1}{P_2}$$

$$\Delta S_{system}=3\times 8.314\times ln\frac{2}{1}$$

$$\Delta S_{system}=17.29J{K}^{-1}$$

ΔS_surrounding is given by:

$$\Delta S_{surrounding}=\frac{-q_{irr}}{T}$$

Let's calculate q_irr by the help of first law of thermodynamics:

$$\Delta U=q+W$$

$$0=q_{irr}-P_{ext}(V_2-V_1)$$

$$0=q_{irr}-P_{ext}(\frac{nRT}{P_2}-\frac{nRT}{P_1})$$

$$q_{irr}=P_{ext}nRT(\frac{1}{P_2}-\frac{1}{P_1})$$

$$q_{irr}=1\times 3\times 8.314\times 300(\frac{1}{1}-\frac{1}{2})$$

$$q_{irr}=3741.3J$$

Putting this value in ΔS_surrounding:

$$\Delta S_{surrounding}=\frac{-3741.3}{300}=-12.47J{K}^{-1}$$

Total change in entropy is given by:

$$\Delta S_{total}=\Delta S_{system}+\Delta S_{surrounding}$$

$$\Delta S_{total}=17.29-12.47=4.82$$

Here, ΔS_total > 0. So, process is spontaneous and irreversible.

100 g of N₂ gas at 300 K is held by a piston under 30 atm. The pressure is suddenly released to 10 atm and gas is allowed to expand adiabatically. If C_v is 20.8 JK^-1mol^-1, calculate ΔS_system.

• Moles of N₂, n = 100/28

• Since, the process is adiabatic, q = 0

• Using first law of thermodynamics:

$$\Delta U=q+W$$

$$\Delta U=W$$

$$n{C}_v(T_2-T_1)=-P_{ext}(V_2-V_1)$$

$$n{C}_v(T_2-300)=-10(\frac{nR{T}_2}{P_2}-\frac{nR{T}_1}{P_1})$$

$$n\times 20.8(T_2-300)=-10(\frac{nR{T}_2}{P_2}-\frac{nR{T}_1}{P_1})$$

$$20.8(T_2-300)=-10\frac{R{T}_2}{P_2}+10\frac{R{T}_1}{P_1}$$

$$20.8T_2-(300\times 20.8)=-10\frac{8.314\times T_2}{10}+10\frac{8.314\times 300}{30}$$

$$20.8T_2-6240=-8.314T_2+831.4$$

$$(20.8+8.314)T_2=6240+831.4$$

$$T_2=\frac{7071.4}{29.114}=243K(approx.)$$

Change in enthalpy of the system is given by:

$$\Delta S=n{C}_{p}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$$

$$\Delta S=\frac{100}{28}\times (20.8+R)ln\frac{243}{300}+\frac{100}{28}\times 8.314\times ln\frac{30}{10}$$

$$\Delta S=10.71J{K}^{-1}$$

A certain chemical reaction is spontaneous at 72^oC. If the enthalpy change for the reaction is 19 KJ, what is the minimum value of ΔS for the reaction?

Since, the reaction takes place at constant temperature. So, q_rev = ΔH = 19 KJ = 19000 J.

Temperature, T = 72^oC = (72 + 273)K = 345 K

Change in entropy is given by:

$$\Delta S=\frac{q_{rev}}{T}=\frac{\Delta H}{T}$$

$$\Delta S=\frac{19000}{345}=55.07J{K}^{-1}$$