30 Isobaric thermodynamics processes – problems and solutions
1. ਪੀਵੀ ਡਾਇਗ੍ਰਾਮ ਹੇਠਾਂ ਇੱਕ ਦਿਖਾਉਂਦਾ ਹੈ ਆਦਰਸ਼ ਗੈਸ ਇੱਕ ਆਈਸੋ ਵਿੱਚੋਂ ਗੁਜ਼ਰਦਾ ਹੈbaric ਪ੍ਰਕਿਰਿਆ। ਗਣਨਾ ਕਰੋ ਦਾ ਕੰਮ ਇਹ ਪ੍ਰਕਿਰਿਆ AB ਵਿੱਚ ਗੈਸ ਦੁਆਰਾ ਕੀਤੀ ਜਾਂਦੀ ਹੈ।
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
ਦਬਾਅ (P) = 5 x 105 N / ਮੀਟਰ2
ਸ਼ੁਰੂਆਤੀ ਵਾਲੀਅਮ (V1) = 2 ਮੀਟਰ3
ਅੰਤਿਮ ਵਾਲੀਅਮ (V2) = 6 ਮੀਟਰ3
ਲੋੜੀਂਦਾ: ਕੰਮ (W)
ਹੱਲ:
W = P (V)2 - ਵੀ1)
W = (5 x 10)5)(6 – 2) = (5 x 105) (4..XNUMX)
ਡਬਲਯੂ = 20 x 105 = 2x106 joule
2. What is difference of the work is done by the gas in process AB and process CD…
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
Isobaric process AB :
Pressure (P) = 6 atm = 6 x 105 N / ਮੀਟਰ2
ਸ਼ੁਰੂਆਤੀ ਵਾਲੀਅਮ (V1) = 1 ਲੀਟਰ = 1 ਡੀਐਮ3 = 1x10-3 m3
ਅੰਤਿਮ ਵਾਲੀਅਮ (V2) = 3 liters = 3 dm3 = 3x10-3 m3
Isobaric process CD :
Pressure (P) = 4 atm = 4 x 105 N / ਮੀਟਰ2
ਸ਼ੁਰੂਆਤੀ ਵਾਲੀਅਮ (V1) = 2 liters = 2 dm3 = 2x10-3 m3
ਅੰਤਿਮ ਵਾਲੀਅਮ (V2) = 5 liters = 5 dm3 = 5x10-3 m3
ਲੋੜੀਂਦਾ : Difference of the work is done by the gas in process AB and CD.
ਹੱਲ:
Work is done by the gas in process AB :
W = P (V)2 - ਵੀ1)
W = (6 x 10)5)(3 x 10-3 - 1 x 10-3)
W = (6 x 10)5)(2 x 10-3)
ਡਬਲਯੂ = 12 x 102 = 1200 ਜੂਲ
Work is done by the gas in process CD :
W = P (V)2 - ਵੀ1)
W = (4 x 10)5)(5 x 10-3 - 2 x 10-3)
W = (4 x 10)5)(3 x 10-3)
ਡਬਲਯੂ = 12 x 102 = 1200 ਜੂਲ
Difference of the work is done by the gas in process AB and CD = 1200 – 1200 = 0.
3. Work is done by the gas in process ABC is….
ਜਾਣਿਆ ਜਾਂਦਾ ਹੈ:
ਦਬਾਅ 1 (P1) = 6 x 105 ਪਾ = 6 x 105 N / ਮੀਟਰ2
ਦਬਾਅ 2 (P2) = 3 x 105 ਪਾ = 3 x 105 N / ਮੀਟਰ2
ਖੰਡ 1 (V)1) = 2 ਸੈ.ਮੀ3 = 2x10-6 m3
ਖੰਡ 2 (V)2) = 6 ਸੈ.ਮੀ3 = 6x10-6 m3
ਲੋੜੀਂਦਾ : Work is done in process ABC.
ਹੱਲ:
In process AB, the volume is kept constant so that no work is done by the gas.
Work was done by the gas in the process BC.
W = P2 (V2 - ਵੀ1)
W = (3 x 10)5)(6 x 10-6 - 2 x 10-6)
W = (3 x 10)5)(4 x 10-6)
ਡਬਲਯੂ = 12 x 10-1
ਡਬਲਯੂ = 1.2 ਜੂਲ
Work is done in the process ABC = work is done in the process AB = 1.2 Joule.
4. Determine the change in internal energy for 2 moles of an ideal gas undergoing an isobaric expansion at 300 K, where \(\Delta V = 1\ \text{m}^3\).
Solution: \(\Delta U = nC_v\Delta T\), using \(C_v = \frac{R}{\gamma-1}\) (for monatomic ideal gas, \(\gamma = \frac{5}{3}\)) and \(\Delta T = \frac{P\Delta V}{nR}\), \(\Delta U = \frac{2\cdot 300 \cdot 1}{\frac{5}{3}-1} \approx 1800\ \text{J}\).
5. Calculate the heat transfer in an isobaric process where 1 mole of a diatomic ideal gas expands, \(C_p = \frac{7}{2}R\), and \(\Delta T = 50\ \text{K}\).
Solution: \(Q = nC_p\Delta T = \frac{7}{2} \cdot 50 \cdot R \approx 1750\ \text{J}\) (using \(R = 8.314\ \text{J/(mol·K)}\)).
6. Find the work done by a system undergoing an isobaric expansion, \(P = 3\ \text{atm}\), \(\Delta V = 4\ \text{L}\).
Solution: \(W = P\Delta V = 3 \times 4 = 12\ \text{L·atm}\).
7. Determine the change in entropy for an isobaric process where 2 moles of an ideal gas change temperature by 20 K. Use \(C_p = \frac{5}{2}R\).
Solution: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 2 \cdot \frac{5}{2}R \cdot \ln\frac{T_1+20}{T_1}\).
8. Calculate the heat transfer for an isobaric compression of a monatomic ideal gas, \(C_p = \frac{5}{2}R\), \(\Delta T = -10\ \text{K}\).
Solution: \(Q = nC_p\Delta T = \frac{5}{2} \cdot (-10) \cdot R \approx -415\ \text{J}\).
9. Find the work done on the system in an isobaric process with \(P = 5\ \text{bar}\), \(\Delta V = -3\ \text{m}^3\).
Solution: \(W = P\Delta V = 5 \times (-3) = -15\ \text{bar·m}^3\).
10. Determine the change in internal energy for an isobaric process where \(n = 3\ \text{mol}\), \(C_v = 3R\), \(\Delta T = 25\ \text{K}\).
Solution: \(\Delta U = nC_v\Delta T = 3 \cdot 3R \cdot 25 \approx 1883\ \text{J}\).
11. Calculate the entropy change in an isobaric process for a diatomic ideal gas, \(n = 1\ \text{mol}\), \(\Delta T = 40\ \text{K}\), \(T_1 = 300\ \text{K}\).
Solution: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = \frac{7}{2}R\ln\frac{340}{300}\).
12. Find the heat transfer in an isobaric expansion, \(P = 2\ \text{atm}\), \(\Delta V = 3\ \text{L}\), \(C_p = \frac{7}{2}R\).
Solution: \(Q = P\Delta V + nC_p\Delta T = 2 \times 3 + \frac{7}{2}R\Delta T\).
13. Determine the work done in an isobaric process for \(P = 4\ \text{bar}\), \(\Delta V = 5\ \text{m}^3\).
Solution: \(W = P\Delta V = 4 \times 5 = 20\ \text{bar·m}^3\).
14. Calculate the internal energy change for an isobaric compression, \(n = 2\ \text{mol}\), \(C_v = \frac{3}{2}R\), \(\Delta T = -30\ \text{K}\).
Solution: \(\Delta U = nC_v\Delta T = 2 \cdot \frac{3}{2}R \cdot (-30) \approx -753\ \text{J}\).
15. Find the entropy change in an isobaric process, \(n = 1.5\ \text{mol}\), \(\Delta T = 60\ \text{K}\), \(T_1 = 400\ \text{K}\), \(C_p = \frac{5}{2}R\).
Solution: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 1.5 \cdot \frac{5}{2}R\ln\frac{460}{400}\).
16. Determine the heat transfer for an isobaric expansion, \(P = 3\ \text{bar}\), \(\Delta V = 2\ \text{m}^3\), \(C_p = \frac{5}{2}R\), \(n = 2\ \text{mol}\).
Solution: \(Q = P\Delta V + nC_p\Delta T = 3 \times 2 + 2 \cdot \frac{5}{2}R\Delta T\).
17. Calculate the work done on 3 moles of a gas undergoing an isobaric compression, \(P = 5\ \text{atm}\), \(\Delta V = -4\ \text{L}\).
Solution: \(W = P\Delta V = 5 \times (-4) = -20\ \text{L·atm}\).
18. Determine the internal energy change for \(n = 4\ \text{mol}\), \(C_v = \frac{7}{2}R\), \(\Delta T = 15\ \text{K}\) in an isobaric process.
Solution: \(\Delta U = nC_v\Delta T = 4 \cdot \frac{7}{2}R \cdot 15 \approx 3157\ \text{J}\).
19. Find the heat transfer in an isobaric process, \(P = 4\ \text{atm}\), \(\Delta V = 5\ \text{L}\), \(n = 2\ \text{mol}\), \(C_p = \frac{5}{2}R\).
Solution: \(Q = P\Delta V + nC_p\Delta T = 4 \times 5 + 2 \cdot \frac{5}{2}R\Delta T\).
20. Determine the work done in an isobaric compression, \(P = 7\ \text{bar}\), \(\Delta V = -2\ \text{m}^3\).
Solution: \(W = P\Delta V = 7 \times (-2) = -14\ \text{bar·m}^3\).
21. Calculate the internal energy change for 3 moles of an ideal gas undergoing an isobaric process, \(C_v = \frac{5}{2}R\), \(\Delta T = 20\ \text{K}\).
Solution: \(\Delta U = nC_v\Delta T = 3 \cdot \frac{5}{2}R \cdot 20 \approx 1256\ \text{J}\).
22. Find the entropy change for an isobaric expansion, \(n = 1\ \text{mol}\), \(C_p = \frac{7}{2}R\), \(\Delta T = 30\ \text{K}\), \(T_1 = 250\ \text{K}\).
Solution: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = \frac{7}{2}R\ln\frac{280}{250}\).
23. Determine the heat transfer in an isobaric process, \(P = 6\ \text{bar}\), \(\Delta V = 4\ \text{m}^3\), \(n = 3\ \text{mol}\), \(C_p = \frac{3}{2}R\).
Solution: \(Q = P\Delta V + nC_p\Delta T = 6 \times 4 + 3 \cdot \frac{3}{2}R\Delta T\).
24. Calculate the work done by the system in an isobaric expansion with \(P = 8\ \text{bar}\), \(\Delta V = 3\ \text{m}^3\).
Solution: \(W = P\Delta V = 8 \times 3 = 24\ \text{bar·m}^3\).
25. Determine the internal energy change for an isobaric process where \(n = 2\ \text{mol}\), \(C_v = \frac{7}{2}R\), \(\Delta T = -10\ \text{K}\).
Solution: \(\Delta U = nC_v\Delta T = 2 \cdot \frac{7}{2}R \cdot (-10) \approx -878\ \text{J}\).
26. Find the entropy change for a diatomic ideal gas in an isobaric compression, \(n = 1.5\ \text{mol}\), \(T_1 = 350\ \text{K}\), \(\Delta T = -40\ \text{K}\).
Solution: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 1.5 \cdot \frac{7}{2}R\ln\frac{310}{350}\).
27. Determine the heat transfer for 2 moles of a gas undergoing an isobaric expansion, \(P = 5\ \text{bar}\), \(\Delta V = 6\ \text{m}^3\), \(C_p = \frac{5}{2}R\).
Solution: \(Q = P\Delta V + nC_p\Delta T = 5 \times 6 + 2 \cdot \frac{5}{2}R\Delta T\).
28. Calculate the work done on the system in an isobaric compression with \(P = 9\ \text{atm}\), \(\Delta V = -3\ \text{L}\).
Solution: \(W = P\Delta V = 9 \times (-3) = -27\ \text{L·atm}\).
29. Determine the internal energy change for 3 moles of a gas undergoing an isobaric process, \(C_v = \frac{3}{2}R\), \(\Delta T = 15\ \text{K}\).
Solution: \(\Delta U = nC_v\Delta T = 3 \cdot \frac{3}{2}R \cdot 15 \approx 564\ \text{J}\).
30. Find the entropy change in an isobaric expansion, \(n = 4\ \text{mol}\), \(C_p = \frac{5}{2}R\), \(\Delta T = 25\ \text{K}\), \(T_1 = 300\ \text{K}\).
Solution: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 4 \cdot \frac{5}{2}R\ln\frac{325}{300}\).
