# Isobaric thermodynamics processes – problems and solutions

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 105 N/m2

Initial volume (V1) = 2 m3

Final volume (V2) = 6 m3

Wanted : Work (W)

Solution :

W = P (V2 – V1)

W = (5 x 105)(6 – 2) = (5 x 105) (4)

W = 20 x 105 = 2 x 106 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 105 N/m2

Initial volume (V1) = 1 liter = 1 dm3 = 1 x 10-3 m3

Final volume (V2) = 3 liters = 3 dm3 = 3 x 10-3 m3

Isobaric process CD :

Pressure (P) = 4 atm = 4 x 105 N/m2

Initial volume (V1) = 2 liters = 2 dm3 = 2 x 10-3 m3

Final volume (V2) = 5 liters = 5 dm3 = 5 x 10-3 m3

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 (V2 – V1)

W = (6 x 105)(3 x 10-3 – 1 x 10-3)

W = (6 x 105)(2 x 10-3)

W = 12 x 102 = 1200 Joule

Work is done by the gas in process CD :

W = P (V2 – V1)

W = (4 x 105)(5 x 10-3 – 2 x 10-3)

W = (4 x 105)(3 x 10-3)

W = 12 x 102 = 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 (P1) = 6 x 105 Pa = 6 x 105 N/m2

Pressure 2 (P2) = 3 x 105 Pa = 3 x 105 N/m2

Volume 1 (V1) = 2 cm3 = 2 x 10-6 m3

Volume 2 (V2) = 6 cm3 = 6 x 10-6 m3

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 = P2 (V2 – V1)

W = (3 x 105)(6 x 10-6 – 2 x 10-6)

W = (3 x 105)(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: $$\Delta U = nC_v\Delta T = 3 \cdot \frac{5}{2}R \cdot 20 \approx 1256\ \text{J}$$.

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