30 Isobaric thermodynamics processes – problems and solutions

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

__Known :__

Pressure (P) = 5 x 10^{5} N/m^{2}

Initial volume (V_{1}) = 2 m^{3}

Final volume (V_{2}) = 6 m^{3}

__Wanted :__ Work (W)

__Solution :__

W = P (V_{2} – V_{1})

W = (5 x 10^{5})(6 – 2) = (5 x 10^{5}) (4)

W = 20 x 10^{5 }= 2 x 10^{6} Joule

2. What is difference of the work is done by the gas in process AB and process CD…

__Known :__

**Isobaric process ****AB** :

Pressure (P) = 6 atm = 6 x 10^{5} N/m^{2}

Initial volume (V_{1}) = 1 liter = 1 dm^{3} = 1 x 10^{-3} m^{3}

Final volume (V_{2}) = 3 liters = 3 dm^{3} = 3 x 10^{-3} m^{3 }

**Isobaric process CD **:

Pressure (P) = 4 atm = 4 x 10^{5} N/m^{2}

Initial volume (V_{1}) = 2 liters = 2 dm^{3} = 2 x 10^{-3} m^{3}

Final volume (V_{2}) = 5 liters = 5 dm^{3} = 5 x 10^{-3} m^{3}

__Wanted ____:__ Difference of the work is done by the gas in process AB and CD.

__Solution :__

Work is done by the gas in process AB :

W = P (V_{2} – V_{1})

W = (6 x 10^{5})(3 x 10^{-3 }– 1 x 10^{-3})

W = (6 x 10^{5})(2 x 10^{-3})

W = 12 x 10^{2 }= 1200 Joule

Work is done by the gas in process CD :

W = P (V_{2} – V_{1})

W = (4 x 10^{5})(5 x 10^{-3 }– 2 x 10^{-3})

W = (4 x 10^{5})(3 x 10^{-3})

W = 12 x 10^{2 }= 1200 Joule

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….

__Known :__

Pressure 1 (P_{1}) = 6 x 10^{5} Pa = 6 x 10^{5} N/m^{2}

Pressure 2 (P_{2}) = 3 x 10^{5 }Pa = 3 x 10^{5} N/m^{2}

Volume 1 (V_{1}) = 2 cm^{3} = 2 x 10^{-6} m^{3}

Volume 2 (V_{2}) = 6 cm^{3} = 6 x 10^{-6} m^{3}

__Wanted__ : Work is done in process ABC.

__Solution :__

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 = P_{2} (V_{2} – V_{1})

W = (3 x 10^{5})(6 x 10^{-6} – 2 x 10^{-6})

W = (3 x 10^{5})(4 x 10^{-6})

W = 12 x 10^{-1}

W = 1.2 Joule

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: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 4 \cdot \frac{5}{2}R\ln\frac{325}{300}\).