30 Isochoric thermodynamics processes – problems and solutions
1. مخطط الطاقة الشمسية الكهروضوئية يوضح ما يلي غاز مثالي يخضع لعملية تساوي الحرارةchoric العملية. احسب العمل يتم ذلك بواسطة الغاز في العملية AB.
الحل:
Process AB is an عملية متساوية الحجم (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 الضغط 2 س 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
معروف :
في البداية درجة الحرارة (T1) = 47oج + ٢٧٣ = ٢٩٣ كلفن
الضغط الأولي (P)1) = 2 × 105 Pa
الضغط النهائي (P)2) = 3 × 105 Pa
Universal gas constant (R) = 8.315 J/mol.K
Number of moles (n) = 3
مطلوب: The change in internal energy of the gas.
الحل:
In the isochoric process, the volume is kept constant so that no work is done by the gas (W = 0).
القانون الأول للديناميكا الحرارية :
ΔU = QW
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 - تي1)
جاي لوساك‘s law (constant volume) :
The change in internal energy of gas :
ΔU = 3/2 n R (T2 - تي1) = 3/2 (3)(8.315)(480-320)
ΔU = 3/2 (24.945)(160) = 3/2 (3991.2)
ΔU = 5986.8 جول
3. 0.2 moles of monatomic gases at 27oC are in a closed container. The is added to the gas so that temperature of gas becomes 400 K is… Universal gas constant (R) = 8.315 J/mol.K
معروف :
عدد المولات (ن) = 0.2 مول
درجة الحرارة الابتدائية (T)1) = 27oج + ٢٧٣ = ٢٩٣ كلفن
درجة الحرارة النهائية (T)2) = 400 ك
Universal constant gas (R) = 8.315 J/mol.K
مطلوب : Heat is added (Q)
الحل:
In isochoric process, volume is kept constant so that no work is done by the gas (W = 0).
The first law of thermodynamics :
ΔU = QW
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 - تي1)
ΔU = 3/2 (0.2)(8.315)(400-300)
ΔU = 3/2 (0.2)(8.315)(100)
ΔU = 249.45 جول
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.
