30 Isothermal thermodynamic processes – problems and solutions
1. PV diagram below shows an idealios dujos patiria izoterminis procesas. Calculate the work is done by the gas in the process AB.
Sprendimas
Dirbkite done by a gas is equal to the area under the PV curve
AB = triangle area + rectangle area
W = [½ (8 x 105–4 x 105)(3-1)] + [4 x 105 (3-1)]
W = [½ (4 x 105)(2)] + [4 x 105 (2)]
W = [4 x 105] + [8 x 105]
P = 12 x 105 Džaulio
The work is done by the gas in the process AB = X 12 105 Džaulio
2. Calculate the work is done by an ideal gas in the process ABC.
The work is done by an ideal gas in the process ABC = the area under the PV curve
AB = triangle area + rectangle area
W = [½(10×105–5 × 105)(30-10)]+[5×105(30-10)]
W = [½ (5 x 105)(20)] + [5 x 105 (20)]
W = [(5 x 105)(10)] + [100 x 105]
W = [50 x 105] + [100 x 105]
P = 150 x 105 Džaulio
P = 1.5 x 107 Džaulio
3. n ideal gas undergoing isothermal processes. What is an amount of šiluma is added to the gas so the gas do work of 5000 Joule on the environment.
Žinomas:
Darbas (W) = 5000 džaulis
Ieškoma: Heat is added to the gas (Q)
sprendimas:
An isothermal process is a thermodynamic process that occurs at a constant temperatūra.
ΔU = 3/2 n R ΔT
ΔU = the change in internal energy, n = number of moles, R = universal gas constant, ΔT = The change in temperature.
According to the above equation, if ΔT = 0 then ΔU = 0.
The equation of the first law of thermodynamics :
ΔU = Q – W
0 = Q – W
Q = W
Q = 5000 Joule.
4. PV diagram for an ideal gas undergoing isothermal process shown in the figure below. Calculate the heat is added by a gas in process AB.
Žinomas:
1 slėgis (P1) = 5 atmosferos = 5 x 105 Pa
2 slėgis (P2) = 10 atmosferos = 10 x 105 Pa
1 tomas (V1) = 2 m3
2 tomas (V2) = 6 m3
Ieško : Heat is added in process AB.
sprendimas:
Isothermal = constant temperature. According to the equation below, if ΔT = 0 then ΔU = 0.
ΔU = 3/2 n R ΔT
ΔU = 3/2 n R (0)
ΔU = 0
Rašyti paraišką į the first law of thermodynamics :
ΔU = Q-W
0 = Q-W
Q=W
The work is done by an ideal gas = the area under the PV curve = triangle area + rectangle area
W = ½ (P2 - P1)(V2 - V1) + P1 (V2 - V1)
W = ½ (10 x 105 - 5 x 105)(6-2) + (5 x 105)(6–2 m.)
W = ½ (5 x 105)(4) + (5 x 105)(4)
W = ½ (20 x 105) + (20 x 105)
W = (10 x 105) + (20 x 105)
P = 30 x 105 Džaulio
5. Calculate the work done on 1 mol of an ideal gas expanding isothermally from 2 L to 4 L at 300 K.
Solution: \( W = nRT \ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times 300 \times \ln(2) \approx 1724 \, \text{J} \)
6. Determine the heat transfer for the above problem.
Solution: \( Q = W = 1724 \, \text{J} \)
7. Calculate the change in internal energy for the above process.
Solution: \( \Delta U = 0 \, \text{J} \) (since the process is isothermal)
8. For an isothermal compression of an ideal gas from 10 L to 5 L at 300 K, find the work done.
Solution: \( W = 8.314 \times 300 \times \ln\left(\frac{5}{10}\right) \approx -862 \, \text{J} \)
9. Calculate the entropy change for the isothermal expansion in Problem 1.
Solution: \( \Delta S = nR\ln\left(\frac{V_2}{V_1}\right) = 8.314 \times \ln(2) \approx 5.76 \, \text{J/K} \)
10. Find the heat transfer for an isothermal compression of 2 mol of an ideal gas from 4 L to 2 L at 200 K.
Solution: \( Q = 2 \times 8.314 \times 200 \times \ln\left(\frac{2}{4}\right) \approx -1152 \, \text{J} \)
11. Calculate the change in Gibbs free energy for an isothermal process.
Solution: \( \Delta G = 0 \) (For a reversible isothermal process in a closed system, \( \Delta G = 0 \))
12. Determine the entropy change for an isothermal compression of 3 mol of an ideal gas from 6 L to 3 L at 400 K.
Solution: \( \Delta S = 3 \times 8.314 \times \ln\left(\frac{3}{6}\right) \approx -17.29 \, \text{J/K} \)
13. Find the work done by an ideal gas expanding isothermally from 3 L to 6 L at 250 K for 2 mol.
Solution: \( W = 2 \times 8.314 \times 250 \times \ln(2) \approx 2874 \, \text{J} \)
14. Determine the heat transfer for the above process.
Solution: \( Q = W = 2874 \, \text{J} \)
15. Calculate the change in internal energy for the above isothermal expansion.
Solution: \( \Delta U = 0 \, \text{J} \)
16. Calculate the entropy change for an isothermal expansion of 4 mol of an ideal gas from 5 L to 10 L at 500 K.
Solution: \( \Delta S = 4 \times 8.314 \times \ln(2) \approx 23.03 \, \text{J/K} \)
17. Determine the heat transfer for an isothermal compression of 1 mol of an ideal gas from 8 L to 4 L at 300 K.
Solution: \( Q = 8.314 \times 300 \times \ln\left(\frac{4}{8}\right) \approx -862 \, \text{J} \)
18. Find the work done by 3 mol of an ideal gas expanding isothermally from 3 L to 9 L at 350 K.
Solution: \( W = 3 \times 8.314 \times 350 \times \ln(3) \approx 5362 \, \text{J} \)
19. Calculate the entropy change for the above process.
Solution: \( \Delta S = 3 \times 8.314 \times \ln(3) \approx 14.88 \, \text{J/K} \)
20. Determine the heat transfer for an isothermal expansion of 2 mol of an ideal gas from 2 L to 8 L at 200 K.
Solution: \( Q = 2 \times 8.314 \times 200 \times \ln(4) \approx 2304 \, \text{J} \)
21. Calculate the change in internal energy for an isothermal compression of 4 mol of an ideal gas from 10 L to 5 L at 500 K.
Solution: \( \Delta U = 0 \, \text{J} \)
22. Determine the entropy change for an isothermal expansion of 5 mol of an ideal gas from 5 L to 15 L at 300 K.
Solution: \( \Delta S = 5 \times 8.314 \times \ln(3) \approx 24.81 \, \text{J/K} \)
23. Find the work done by 2 mol of an ideal gas compressing isothermally from 6 L to 2 L at 400 K.
Solution: \( W = 2 \times 8.314 \times 400 \times \ln\left(\frac{2}{6}\right) \approx -2874 \, \text{J} \)
24. Calculate the entropy change for an isothermal compression of 3 mol of an ideal gas from 9 L to 3 L at 250 K.
Solution: \( \Delta S = 3 \times 8.314 \times \ln\left(\frac{3}{9}\right) \approx -14.88 \, \text{J/K} \)
25. Determine the heat transfer for an isothermal expansion of 1 mol of an ideal gas from 4 L to 12 L at 350 K.
Solution: \( Q = 8.314 \times 350 \times \ln(3) \approx 1791 \, \text{J} \)
26. Find the work done by 4 mol of an ideal gas expanding isothermally from 4 L to 8 L at 500 K.
Solution: \( W = 4 \times 8.314 \times 500 \times \ln(2) \approx 5749 \, \text{J} \)
27. Determine the entropy change for an isothermal compression of 2 mol of an ideal gas from 10 L to 5 L at 200 K.
Solution: \( \Delta S = 2 \times 8.314 \times \ln\left(\frac{5}{10}\right) \approx -5.76 \, \text{J/K} \)
24. Find the work done by 3 mol of an ideal gas compressing isothermally from 9 L to 3 L at 450 K.
Solution: \( W = 3 \times 8.314 \times 450 \times \ln\left(\frac{3}{9}\right) \approx -4310 \, \text{J} \)
25. Calculate the change in internal energy for an isothermal expansion of 5 mol of an ideal gas from 5 L to 10 L at 400 K.
Solution: \( \Delta U = 0 \, \text{J} \)
26. Determine the entropy change for an isothermal expansion of 4 mol of an ideal gas from 6 L to 18 L at 300 K.
Solution: \( \Delta S = 4 \times 8.314 \times \ln(3) \approx 19.85 \, \text{J/K} \)
These problems cover various aspects of isothermal processes, including work done, heat transfer, changes in internal energy, and entropy change.
