Isochoric thermodynamics processes – problems and solutions

30 Isochoric thermodynamics processes – problems and solutions

1. PV diagram below shows an ideal gas undergoes an isochoric process. Calculate the work is done by the gas in the process AB.

Isochoric thermodynamics processes - problems and solutions 1Solution :

Process AB is an isochoric process (constant volume). The volume is constant so that no work is done by the gas.

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2. Three moles of monoatomic gas at 47oC and at pressure 2 x 105 Pa, undergoes isochoric process so that pressure increases 3 x 105 Pa. The change in internal energy of the gas is… Universal gas constant (R) = 8.315 J/mol.K

Known :

Initial temperature (T1) = 47oC + 273 = 320 K

Initial pressure (P1) = 2 x 105 Pa

Final pressure (P2) = 3 x 105 Pa

Universal gas constant (R) = 8.315 J/mol.K

Number of moles (n) = 3

Wanted: The change in internal energy of the gas.

Solution :

In the isochoric process, the volume is kept constant so that no work is done by the gas (W = 0).

The first law of thermodynamics :

ΔU = Q-W

ΔU = Q-0

ΔU = Q

ΔU = internal energy, Q = heat

Internal energy of gas :

ΔU = 3/2 n R ΔT = 3/2 n R (T2 – T1)

Gay-Lussac‘s law (constant volume) :

Isochoric thermodynamics processes - problems and solutions 2

The change in internal energy of gas :

ΔU = 3/2 n R (T2 – T1) = 3/2 (3)(8.315)(480-320)

ΔU = 3/2 (24.945)(160) = 3/2 (3991.2)

ΔU = 5986.8 Joule

3. 0.2 moles of monatomic gases at 27oC are in a closed container. The heat is added to the gas so that temperature of gas becomes 400 K is… Universal gas constant (R) = 8.315 J/mol.K

Known :

Number of moles (n) = 0.2 mol

Initial temperature (T1) = 27oC + 273 = 300 K

Final temperature (T2) = 400 K

Universal constant gas (R) = 8.315 J/mol.K

Wanted : Heat is added (Q)

Solution :

In isochoric process, volume is kept constant so that no work is done by the gas (W = 0).

The first law of thermodynamics :

ΔU = Q-W

ΔU = Q-0

ΔU = Q

ΔU = internal energy, Q = heat

The internal energy of gas :

ΔU = 3/2 n R ΔT = 3/2 n R (T2 – T1)

ΔU = 3/2 (0.2)(8.315)(400-300)

ΔU = 3/2 (0.2)(8.315)(100)

ΔU = 249.45 Joule

4. Calculate the heat transfer for an ideal gas undergoing an isochoric process from an initial temperature of 300 K to a final temperature of 400 K. Assume 2 mol of gas, and the molar heat capacity at constant volume (Cᵥ) is 20 J/(mol·K).
Solution: ΔQ = n × Cᵥ × ΔT = 2 mol × 20 J/(mol·K) × (400 K – 300 K) = 4000 J

5. Find the change in internal energy for the above problem.
Solution: ΔU = ΔQ = 4000 J

6. Determine the work done on a system during an isochoric process for the above conditions.
Solution: W = 0 J (since the volume doesn’t change, no work is done)

7. For a monatomic ideal gas undergoing an isochoric process, if the initial pressure is 2 atm and the final pressure is 3 atm, what is the ratio of final to initial temperatures?
Solution: Since P₁/T₁ = P₂/T₂, T₂/T₁ = 3/2

8. What is the entropy change for an ideal gas in an isochoric process when the temperature changes from 300 K to 600 K, and n = 2 mol, Cᵥ = 20 J/(mol·K)?
Solution: ΔS = n × Cᵥ × ln(T₂/T₁) = 2 × 20 × ln(600/300) ≈ 27.73 J/K

9. If the initial state of a diatomic ideal gas is defined by V = 2 L, P = 1 atm, and T = 300 K, find the final pressure if the temperature is doubled in an isochoric process.
Solution: P₂ = 2 × P₁ = 2 atm

10. Find the change in Gibbs free energy for an isochoric process.
Solution: ΔG = 0 (For an isochoric process in a closed system, ΔG = 0)

11. Calculate the final temperature of an ideal gas undergoing an isochoric process if the initial temperature is 200 K, and the initial and final pressures are 2 atm and 4 atm, respectively.
Solution: T₂ = 2 × T₁ = 400 K

12. For an ideal gas, if the heat capacity at constant volume (Cᵥ) is 30 J/(mol·K), find the heat transfer when the temperature changes from 300 K to 450 K, with 3 mol of gas.
Solution: ΔQ = n × Cᵥ × ΔT = 3 × 30 × 150 = 13500 J

13. For the same process as above, calculate the change in internal energy.
Solution: ΔU = ΔQ = 13500 J

14. Determine the entropy change for an isochoric process with n = 1 mol, Cᵥ = 25 J/(mol·K), T₁ = 200 K, and T₂ = 400 K.
Solution: ΔS = n × Cᵥ × ln(T₂/T₁) = 25 × ln(2) ≈ 17.33 J/K

15. Find the work done by the system during an isochoric process of 3 mol of gas, and the temperature changes from 200 K to 300 K.
Solution: W = 0 J (since the volume doesn’t change, no work is done)

16. Calculate the heat transfer for an ideal gas undergoing an isochoric process with an initial temperature of 150 K, final temperature of 300 K, and Cᵥ = 15 J/(mol·K) for 4 mol of gas.
Solution: ΔQ = n × Cᵥ × ΔT = 4 × 15 × 150 = 9000 J

17. What is the entropy change for an ideal gas in an isochoric process with n = 1 mol, Cᵥ = 30 J/(mol·K), T₁ = 100 K, and T₂ = 200 K?
Solution: ΔS = n × Cᵥ × ln(T₂/T₁) = 30 × ln(2) ≈ 20.79 J/K

18. Determine the final pressure of a gas undergoing an isochoric process, given that P₁ = 5 atm, T₁ = 250 K, and T₂ = 500 K.
Solution: P₂ = (T₂/T₁) × P₁ = 2 × 5 atm = 10 atm

19. Find the heat transfer for 5 mol of a monatomic ideal gas undergoing an isochoric process from 300 K to 600 K. Assume Cᵥ = 15 J/(mol·K).
Solution: ΔQ = n × Cᵥ × ΔT = 5 × 15 × 300 = 22500 J

20. What is the change in internal energy for the above problem?
Solution: ΔU = ΔQ = 22500 J

21. Determine the entropy change for an isochoric process where n = 2 mol, Cᵥ = 25 J/(mol·K), T₁ = 300 K, and T₂ = 600 K.
Solution: ΔS = n × Cᵥ × ln(T₂/T₁) = 2 × 25 × ln(2) ≈ 34.66 J/K

22. Calculate the final temperature of 1 mol of a monatomic ideal gas undergoing an isochoric process if the initial temperature is 400 K, and the initial and final pressures are 3 atm and 6 atm, respectively.
Solution: T₂ = 2 × T₁ = 800 K

23. For a diatomic ideal gas undergoing an isochoric process, calculate the change in internal energy when the temperature changes from 300 K to 600 K, with 2 mol of gas, and Cᵥ = 30 J/(mol·K).
Solution: ΔU = n × Cᵥ × ΔT = 2 × 30 × 300 = 18000 J

24. Calculate the heat transfer for an ideal gas undergoing an isochoric process with an initial temperature of 100 K, final temperature of 300 K, and Cᵥ = 20 J/(mol·K) for 2 mol of gas.
Solution: ΔQ = n × Cᵥ × ΔT = 2 × 20 × 200 = 8000 J

25. Find the work done on the system during an isochoric process for the above conditions.
Solution: W = 0 J (since the volume doesn’t change, no work is done)

26. What is the entropy change for an ideal gas in an isochoric process when the temperature changes from 400 K to 800 K, and n = 3 mol, Cᵥ = 20 J/(mol·K)?
Solution: ΔS = n × Cᵥ × ln(T₂/T₁) = 3 × 20 × ln(2) ≈ 41.58 J/K

27. Find the change in Gibbs free energy for an isochoric process.
Solution: ΔG = 0 (For an isochoric process in a closed system, ΔG = 0)

28. Determine the final pressure of a gas undergoing an isochoric process, given that P₁ = 3 atm, T₁ = 300 K, and T₂ = 450 K.
Solution: P₂ = (T₂/T₁) × P₁ = 1.5 × 3 atm = 4.5 atm

29. Calculate the change in internal energy for a system undergoing an isochoric process with 3 mol of gas, Cᵥ = 20 J/(mol·K), and the temperature changes from 200 K to 400 K.
Solution: ΔU = n × Cᵥ × ΔT = 3 × 20 × 200 = 12000 J

30. Determine the entropy change for an isochoric process where n = 4 mol, Cᵥ = 30 J/(mol·K), T₁ = 150 K, and T₂ = 300 K.
Solution: ΔS = n × Cᵥ × ln(T₂/T₁) = 4 × 30 × ln(2) ≈ 55.86 J/K

These problems cover various concepts related to isochoric processes, such as heat transfer, internal energy change, work done, entropy change, and more.

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