Vardaan Learning Institute
Excellence in Foundation & JEE/NEET
Thermodynamics: Solutions to 100 Questions
Topic 1: First Law of Thermodynamics & Internal Energy
Q1.
During a process, the internal energy of a gas increases by 156 J while it absorbs 383 J of heat. Find the work done by the gas.
$\Delta U = +156$ J, $\Delta Q = +383$ J.
$\Delta W = \Delta Q - \Delta U = 383 - 156 = 227$ J
$\Delta W = \Delta Q - \Delta U = 383 - 156 = 227$ J
Q2.
A system absorbs 447 J of heat and does 59 J of work on its surroundings. What is the change in internal energy?
$\Delta Q = +447$ J, $\Delta W = +59$ J.
$\Delta U = \Delta Q - \Delta W = 447 - 59 = 388$ J
$\Delta U = \Delta Q - \Delta W = 447 - 59 = 388$ J
Q3.
During a process, the internal energy of a gas increases by 389 J while it absorbs 734 J of heat. Find the work done by the gas.
$\Delta U = +389$ J, $\Delta Q = +734$ J.
$\Delta W = \Delta Q - \Delta U = 734 - 389 = 345$ J
$\Delta W = \Delta Q - \Delta U = 734 - 389 = 345$ J
Q4.
During a process, the internal energy of a gas increases by 284 J while it absorbs 595 J of heat. Find the work done by the gas.
$\Delta U = +284$ J, $\Delta Q = +595$ J.
$\Delta W = \Delta Q - \Delta U = 595 - 284 = 311$ J
$\Delta W = \Delta Q - \Delta U = 595 - 284 = 311$ J
Q5.
A system absorbs 880 J of heat and does 272 J of work on its surroundings. What is the change in internal energy?
$\Delta Q = +880$ J, $\Delta W = +272$ J.
$\Delta U = \Delta Q - \Delta W = 880 - 272 = 608$ J
$\Delta U = \Delta Q - \Delta W = 880 - 272 = 608$ J
Q6.
During a process, the internal energy of a gas increases by 240 J while it absorbs 735 J of heat. Find the work done by the gas.
$\Delta U = +240$ J, $\Delta Q = +735$ J.
$\Delta W = \Delta Q - \Delta U = 735 - 240 = 495$ J
$\Delta W = \Delta Q - \Delta U = 735 - 240 = 495$ J
Q7.
A system absorbs 467 J of heat and does 110 J of work on its surroundings. What is the change in internal energy?
$\Delta Q = +467$ J, $\Delta W = +110$ J.
$\Delta U = \Delta Q - \Delta W = 467 - 110 = 357$ J
$\Delta U = \Delta Q - \Delta W = 467 - 110 = 357$ J
Q8.
During a process, the internal energy of a gas increases by 286 J while it absorbs 888 J of heat. Find the work done by the gas.
$\Delta U = +286$ J, $\Delta Q = +888$ J.
$\Delta W = \Delta Q - \Delta U = 888 - 286 = 602$ J
$\Delta W = \Delta Q - \Delta U = 888 - 286 = 602$ J
Q9.
During a process, the internal energy of a gas increases by 343 J while it absorbs 255 J of heat. Find the work done by the gas.
$\Delta U = +343$ J, $\Delta Q = +255$ J.
$\Delta W = \Delta Q - \Delta U = 255 - 343 = -88$ J
$\Delta W = \Delta Q - \Delta U = 255 - 343 = -88$ J
Q10.
A system absorbs 552 J of heat and does 219 J of work on its surroundings. What is the change in internal energy?
$\Delta Q = +552$ J, $\Delta W = +219$ J.
$\Delta U = \Delta Q - \Delta W = 552 - 219 = 333$ J
$\Delta U = \Delta Q - \Delta W = 552 - 219 = 333$ J
Topic 2: Specific Heats & Degrees of Freedom
Q11.
For a specific ideal gas, the molar heat capacity at constant volume is $C_v = 3 R$. Calculate its molar heat capacity at constant pressure $C_p$ and the ratio $\gamma$.
By Mayer's relation: $C_p = C_v + R = 4 R$.
$\gamma = C_p/C_v = 4/3 = 1.33$.
$\gamma = C_p/C_v = 4/3 = 1.33$.
Q12.
Calculate the heat required to raise the temperature of 4 moles of a monoatomic ideal gas by 50 K at constant volume. (Take R = 8.31 J/mol K)
For monoatomic gas, $C_v = 1.5 R$.
$Q = n C_v \Delta T = 4 \times 1.5 \times 8.31 \times 50 = 2493.00$ J.
$Q = n C_v \Delta T = 4 \times 1.5 \times 8.31 \times 50 = 2493.00$ J.
Q13.
For a specific ideal gas, the molar heat capacity at constant volume is $C_v = 3 R$. Calculate its molar heat capacity at constant pressure $C_p$ and the ratio $\gamma$.
By Mayer's relation: $C_p = C_v + R = 4 R$.
$\gamma = C_p/C_v = 4/3 = 1.33$.
$\gamma = C_p/C_v = 4/3 = 1.33$.
Q14.
Calculate the heat required to raise the temperature of 3 moles of a monoatomic ideal gas by 44 K at constant volume. (Take R = 8.31 J/mol K)
For monoatomic gas, $C_v = 1.5 R$.
$Q = n C_v \Delta T = 3 \times 1.5 \times 8.31 \times 44 = 1645.38$ J.
$Q = n C_v \Delta T = 3 \times 1.5 \times 8.31 \times 44 = 1645.38$ J.
Q15.
For a specific ideal gas, the molar heat capacity at constant volume is $C_v = 3 R$. Calculate its molar heat capacity at constant pressure $C_p$ and the ratio $\gamma$.
By Mayer's relation: $C_p = C_v + R = 4 R$.
$\gamma = C_p/C_v = 4/3 = 1.33$.
$\gamma = C_p/C_v = 4/3 = 1.33$.
Q16.
Calculate the heat required to raise the temperature of 5 moles of a monoatomic ideal gas by 16 K at constant volume. (Take R = 8.31 J/mol K)
For monoatomic gas, $C_v = 1.5 R$.
$Q = n C_v \Delta T = 5 \times 1.5 \times 8.31 \times 16 = 997.20$ J.
$Q = n C_v \Delta T = 5 \times 1.5 \times 8.31 \times 16 = 997.20$ J.
Q17.
For a specific ideal gas, the molar heat capacity at constant volume is $C_v = 1.5 R$. Calculate its molar heat capacity at constant pressure $C_p$ and the ratio $\gamma$.
By Mayer's relation: $C_p = C_v + R = 2.5 R$.
$\gamma = C_p/C_v = 2.5/1.5 = 1.67$.
$\gamma = C_p/C_v = 2.5/1.5 = 1.67$.
Q18.
Calculate the heat required to raise the temperature of 5 moles of a monoatomic ideal gas by 22 K at constant volume. (Take R = 8.31 J/mol K)
For monoatomic gas, $C_v = 1.5 R$.
$Q = n C_v \Delta T = 5 \times 1.5 \times 8.31 \times 22 = 1371.15$ J.
$Q = n C_v \Delta T = 5 \times 1.5 \times 8.31 \times 22 = 1371.15$ J.
Q19.
For a specific ideal gas, the molar heat capacity at constant volume is $C_v = 2.5 R$. Calculate its molar heat capacity at constant pressure $C_p$ and the ratio $\gamma$.
By Mayer's relation: $C_p = C_v + R = 3.5 R$.
$\gamma = C_p/C_v = 3.5/2.5 = 1.40$.
$\gamma = C_p/C_v = 3.5/2.5 = 1.40$.
Q20.
Calculate the heat required to raise the temperature of 5 moles of a monoatomic ideal gas by 27 K at constant volume. (Take R = 8.31 J/mol K)
For monoatomic gas, $C_v = 1.5 R$.
$Q = n C_v \Delta T = 5 \times 1.5 \times 8.31 \times 27 = 1682.78$ J.
$Q = n C_v \Delta T = 5 \times 1.5 \times 8.31 \times 27 = 1682.78$ J.
Topic 3: Isothermal Processes
Q21.
1 moles of an ideal gas expands isothermally at 314 K such that its volume becomes 2 times the initial volume. Find the work done. (Take $R = 8.31, \ln(2) \approx 0.69$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 1 \times 8.31 \times 314 \times \ln(2) \approx 1800.4$ J.
$W = 1 \times 8.31 \times 314 \times \ln(2) \approx 1800.4$ J.
Q22.
3 moles of an ideal gas expands isothermally at 307 K such that its volume becomes 3 times the initial volume. Find the work done. (Take $R = 8.31, \ln(3) \approx 1.1$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 3 \times 8.31 \times 307 \times \ln(3) \approx 8418.9$ J.
$W = 3 \times 8.31 \times 307 \times \ln(3) \approx 8418.9$ J.
Q23.
2 moles of an ideal gas expands isothermally at 358 K such that its volume becomes 2 times the initial volume. Find the work done. (Take $R = 8.31, \ln(2) \approx 0.69$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 2 \times 8.31 \times 358 \times \ln(2) \approx 4105.5$ J.
$W = 2 \times 8.31 \times 358 \times \ln(2) \approx 4105.5$ J.
Q24.
4 moles of an ideal gas expands isothermally at 473 K such that its volume becomes 2 times the initial volume. Find the work done. (Take $R = 8.31, \ln(2) \approx 0.69$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 4 \times 8.31 \times 473 \times \ln(2) \approx 10848.5$ J.
$W = 4 \times 8.31 \times 473 \times \ln(2) \approx 10848.5$ J.
Q25.
4 moles of an ideal gas expands isothermally at 426 K such that its volume becomes 2 times the initial volume. Find the work done. (Take $R = 8.31, \ln(2) \approx 0.69$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 4 \times 8.31 \times 426 \times \ln(2) \approx 9770.6$ J.
$W = 4 \times 8.31 \times 426 \times \ln(2) \approx 9770.6$ J.
Q26.
1 moles of an ideal gas expands isothermally at 452 K such that its volume becomes 3 times the initial volume. Find the work done. (Take $R = 8.31, \ln(3) \approx 1.1$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 1 \times 8.31 \times 452 \times \ln(3) \approx 4131.7$ J.
$W = 1 \times 8.31 \times 452 \times \ln(3) \approx 4131.7$ J.
Q27.
2 moles of an ideal gas expands isothermally at 311 K such that its volume becomes 3 times the initial volume. Find the work done. (Take $R = 8.31, \ln(3) \approx 1.1$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 2 \times 8.31 \times 311 \times \ln(3) \approx 5685.7$ J.
$W = 2 \times 8.31 \times 311 \times \ln(3) \approx 5685.7$ J.
Q28.
1 moles of an ideal gas expands isothermally at 402 K such that its volume becomes 4 times the initial volume. Find the work done. (Take $R = 8.31, \ln(4) \approx 1.39$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 1 \times 8.31 \times 402 \times \ln(4) \approx 4643.5$ J.
$W = 1 \times 8.31 \times 402 \times \ln(4) \approx 4643.5$ J.
Q29.
1 moles of an ideal gas expands isothermally at 457 K such that its volume becomes 2 times the initial volume. Find the work done. (Take $R = 8.31, \ln(2) \approx 0.69$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 1 \times 8.31 \times 457 \times \ln(2) \approx 2620.4$ J.
$W = 1 \times 8.31 \times 457 \times \ln(2) \approx 2620.4$ J.
Q30.
1 moles of an ideal gas expands isothermally at 470 K such that its volume becomes 2 times the initial volume. Find the work done. (Take $R = 8.31, \ln(2) \approx 0.69$)
Isothermal work $W = nRT \ln(V_2/V_1)$.
$W = 1 \times 8.31 \times 470 \times \ln(2) \approx 2694.9$ J.
$W = 1 \times 8.31 \times 470 \times \ln(2) \approx 2694.9$ J.
Topic 4: Adiabatic Processes
Q31.
An ideal diatomic gas ($\gamma = 1.4$) of 1 moles expands adiabatically, and its temperature drops from 401 K to 297 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{1 \times 8.31 \times (401 - 297)}{1.4 - 1} = \frac{864.2}{0.4} = 2160.6$ J.
$W = \frac{1 \times 8.31 \times (401 - 297)}{1.4 - 1} = \frac{864.2}{0.4} = 2160.6$ J.
Q32.
An ideal diatomic gas ($\gamma = 1.4$) of 2 moles expands adiabatically, and its temperature drops from 496 K to 307 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{2 \times 8.31 \times (496 - 307)}{1.4 - 1} = \frac{3141.2}{0.4} = 7853.0$ J.
$W = \frac{2 \times 8.31 \times (496 - 307)}{1.4 - 1} = \frac{3141.2}{0.4} = 7853.0$ J.
Q33.
An ideal diatomic gas ($\gamma = 1.4$) of 1 moles expands adiabatically, and its temperature drops from 512 K to 279 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{1 \times 8.31 \times (512 - 279)}{1.4 - 1} = \frac{1936.2}{0.4} = 4840.6$ J.
$W = \frac{1 \times 8.31 \times (512 - 279)}{1.4 - 1} = \frac{1936.2}{0.4} = 4840.6$ J.
Q34.
An ideal diatomic gas ($\gamma = 1.4$) of 3 moles expands adiabatically, and its temperature drops from 503 K to 246 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{3 \times 8.31 \times (503 - 246)}{1.4 - 1} = \frac{6407.0}{0.4} = 16017.5$ J.
$W = \frac{3 \times 8.31 \times (503 - 246)}{1.4 - 1} = \frac{6407.0}{0.4} = 16017.5$ J.
Q35.
An ideal diatomic gas ($\gamma = 1.4$) of 2 moles expands adiabatically, and its temperature drops from 535 K to 336 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{2 \times 8.31 \times (535 - 336)}{1.4 - 1} = \frac{3307.4}{0.4} = 8268.4$ J.
$W = \frac{2 \times 8.31 \times (535 - 336)}{1.4 - 1} = \frac{3307.4}{0.4} = 8268.4$ J.
Q36.
An ideal diatomic gas ($\gamma = 1.4$) of 2 moles expands adiabatically, and its temperature drops from 514 K to 312 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{2 \times 8.31 \times (514 - 312)}{1.4 - 1} = \frac{3357.2}{0.4} = 8393.1$ J.
$W = \frac{2 \times 8.31 \times (514 - 312)}{1.4 - 1} = \frac{3357.2}{0.4} = 8393.1$ J.
Q37.
An ideal diatomic gas ($\gamma = 1.4$) of 3 moles expands adiabatically, and its temperature drops from 492 K to 309 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{3 \times 8.31 \times (492 - 309)}{1.4 - 1} = \frac{4562.2}{0.4} = 11405.5$ J.
$W = \frac{3 \times 8.31 \times (492 - 309)}{1.4 - 1} = \frac{4562.2}{0.4} = 11405.5$ J.
Q38.
An ideal diatomic gas ($\gamma = 1.4$) of 2 moles expands adiabatically, and its temperature drops from 411 K to 332 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{2 \times 8.31 \times (411 - 332)}{1.4 - 1} = \frac{1313.0}{0.4} = 3282.4$ J.
$W = \frac{2 \times 8.31 \times (411 - 332)}{1.4 - 1} = \frac{1313.0}{0.4} = 3282.4$ J.
Q39.
An ideal diatomic gas ($\gamma = 1.4$) of 1 moles expands adiabatically, and its temperature drops from 405 K to 331 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{1 \times 8.31 \times (405 - 331)}{1.4 - 1} = \frac{614.9}{0.4} = 1537.4$ J.
$W = \frac{1 \times 8.31 \times (405 - 331)}{1.4 - 1} = \frac{614.9}{0.4} = 1537.4$ J.
Q40.
An ideal diatomic gas ($\gamma = 1.4$) of 2 moles expands adiabatically, and its temperature drops from 436 K to 320 K. Calculate the work done by the gas. (Take R = 8.31)
Adiabatic work $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{2 \times 8.31 \times (436 - 320)}{1.4 - 1} = \frac{1927.9}{0.4} = 4819.8$ J.
$W = \frac{2 \times 8.31 \times (436 - 320)}{1.4 - 1} = \frac{1927.9}{0.4} = 4819.8$ J.
Topic 5: Isobaric and Isochoric Processes
Q41.
A gas expands from 9 L to 17 L against a constant external pressure of 4 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 4 \text{ atm} \times (17 - 9) \text{ L} = 32 \text{ L atm}$.
In Joules: $32 \times 101.3 = 3241.6$ J.
$W = 4 \text{ atm} \times (17 - 9) \text{ L} = 32 \text{ L atm}$.
In Joules: $32 \times 101.3 = 3241.6$ J.
Q42.
A gas expands from 6 L to 20 L against a constant external pressure of 1 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 1 \text{ atm} \times (20 - 6) \text{ L} = 14 \text{ L atm}$.
In Joules: $14 \times 101.3 = 1418.2$ J.
$W = 1 \text{ atm} \times (20 - 6) \text{ L} = 14 \text{ L atm}$.
In Joules: $14 \times 101.3 = 1418.2$ J.
Q43.
A gas expands from 7 L to 20 L against a constant external pressure of 3 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 3 \text{ atm} \times (20 - 7) \text{ L} = 39 \text{ L atm}$.
In Joules: $39 \times 101.3 = 3950.7$ J.
$W = 3 \text{ atm} \times (20 - 7) \text{ L} = 39 \text{ L atm}$.
In Joules: $39 \times 101.3 = 3950.7$ J.
Q44.
A gas expands from 9 L to 21 L against a constant external pressure of 1 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 1 \text{ atm} \times (21 - 9) \text{ L} = 12 \text{ L atm}$.
In Joules: $12 \times 101.3 = 1215.6$ J.
$W = 1 \text{ atm} \times (21 - 9) \text{ L} = 12 \text{ L atm}$.
In Joules: $12 \times 101.3 = 1215.6$ J.
Q45.
A gas expands from 10 L to 22 L against a constant external pressure of 2 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 2 \text{ atm} \times (22 - 10) \text{ L} = 24 \text{ L atm}$.
In Joules: $24 \times 101.3 = 2431.2$ J.
$W = 2 \text{ atm} \times (22 - 10) \text{ L} = 24 \text{ L atm}$.
In Joules: $24 \times 101.3 = 2431.2$ J.
Q46.
A gas expands from 6 L to 23 L against a constant external pressure of 3 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 3 \text{ atm} \times (23 - 6) \text{ L} = 51 \text{ L atm}$.
In Joules: $51 \times 101.3 = 5166.3$ J.
$W = 3 \text{ atm} \times (23 - 6) \text{ L} = 51 \text{ L atm}$.
In Joules: $51 \times 101.3 = 5166.3$ J.
Q47.
A gas expands from 9 L to 15 L against a constant external pressure of 5 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 5 \text{ atm} \times (15 - 9) \text{ L} = 30 \text{ L atm}$.
In Joules: $30 \times 101.3 = 3039.0$ J.
$W = 5 \text{ atm} \times (15 - 9) \text{ L} = 30 \text{ L atm}$.
In Joules: $30 \times 101.3 = 3039.0$ J.
Q48.
A gas expands from 7 L to 22 L against a constant external pressure of 4 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 4 \text{ atm} \times (22 - 7) \text{ L} = 60 \text{ L atm}$.
In Joules: $60 \times 101.3 = 6078.0$ J.
$W = 4 \text{ atm} \times (22 - 7) \text{ L} = 60 \text{ L atm}$.
In Joules: $60 \times 101.3 = 6078.0$ J.
Q49.
A gas expands from 7 L to 16 L against a constant external pressure of 1 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 1 \text{ atm} \times (16 - 7) \text{ L} = 9 \text{ L atm}$.
In Joules: $9 \times 101.3 = 911.7$ J.
$W = 1 \text{ atm} \times (16 - 7) \text{ L} = 9 \text{ L atm}$.
In Joules: $9 \times 101.3 = 911.7$ J.
Q50.
A gas expands from 10 L to 16 L against a constant external pressure of 1 atm. Calculate the work done in Joules. (1 atm = 101.3 J/L)
Isobaric work $W = P \Delta V$.
$W = 1 \text{ atm} \times (16 - 10) \text{ L} = 6 \text{ L atm}$.
In Joules: $6 \times 101.3 = 607.8$ J.
$W = 1 \text{ atm} \times (16 - 10) \text{ L} = 6 \text{ L atm}$.
In Joules: $6 \times 101.3 = 607.8$ J.
Topic 6: Work Calculation from P-V Diagrams
Q51.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 3 m$^3$ and 12 m$^3$, and the pressures alternate between 3 atm and 9 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(9 - 3) \text{ atm} \times (12 - 3) \text{ m}^3 = 6 \times 9 = 54$ atm m$^3$.
Work = $54 \times 101300 = 5470200$ J.
Area = $(9 - 3) \text{ atm} \times (12 - 3) \text{ m}^3 = 6 \times 9 = 54$ atm m$^3$.
Work = $54 \times 101300 = 5470200$ J.
Q52.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 5 m$^3$ and 20 m$^3$, and the pressures alternate between 3 atm and 9 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(9 - 3) \text{ atm} \times (20 - 5) \text{ m}^3 = 6 \times 15 = 90$ atm m$^3$.
Work = $90 \times 101300 = 9117000$ J.
Area = $(9 - 3) \text{ atm} \times (20 - 5) \text{ m}^3 = 6 \times 15 = 90$ atm m$^3$.
Work = $90 \times 101300 = 9117000$ J.
Q53.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 2 m$^3$ and 10 m$^3$, and the pressures alternate between 2 atm and 8 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(8 - 2) \text{ atm} \times (10 - 2) \text{ m}^3 = 6 \times 8 = 48$ atm m$^3$.
Work = $48 \times 101300 = 4862400$ J.
Area = $(8 - 2) \text{ atm} \times (10 - 2) \text{ m}^3 = 6 \times 8 = 48$ atm m$^3$.
Work = $48 \times 101300 = 4862400$ J.
Q54.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 5 m$^3$ and 20 m$^3$, and the pressures alternate between 3 atm and 9 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(9 - 3) \text{ atm} \times (20 - 5) \text{ m}^3 = 6 \times 15 = 90$ atm m$^3$.
Work = $90 \times 101300 = 9117000$ J.
Area = $(9 - 3) \text{ atm} \times (20 - 5) \text{ m}^3 = 6 \times 15 = 90$ atm m$^3$.
Work = $90 \times 101300 = 9117000$ J.
Q55.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 1 m$^3$ and 3 m$^3$, and the pressures alternate between 2 atm and 8 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(8 - 2) \text{ atm} \times (3 - 1) \text{ m}^3 = 6 \times 2 = 12$ atm m$^3$.
Work = $12 \times 101300 = 1215600$ J.
Area = $(8 - 2) \text{ atm} \times (3 - 1) \text{ m}^3 = 6 \times 2 = 12$ atm m$^3$.
Work = $12 \times 101300 = 1215600$ J.
Q56.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 5 m$^3$ and 25 m$^3$, and the pressures alternate between 3 atm and 12 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(12 - 3) \text{ atm} \times (25 - 5) \text{ m}^3 = 9 \times 20 = 180$ atm m$^3$.
Work = $180 \times 101300 = 18234000$ J.
Area = $(12 - 3) \text{ atm} \times (25 - 5) \text{ m}^3 = 9 \times 20 = 180$ atm m$^3$.
Work = $180 \times 101300 = 18234000$ J.
Q57.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 2 m$^3$ and 6 m$^3$, and the pressures alternate between 2 atm and 8 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(8 - 2) \text{ atm} \times (6 - 2) \text{ m}^3 = 6 \times 4 = 24$ atm m$^3$.
Work = $24 \times 101300 = 2431200$ J.
Area = $(8 - 2) \text{ atm} \times (6 - 2) \text{ m}^3 = 6 \times 4 = 24$ atm m$^3$.
Work = $24 \times 101300 = 2431200$ J.
Q58.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 3 m$^3$ and 9 m$^3$, and the pressures alternate between 2 atm and 8 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(8 - 2) \text{ atm} \times (9 - 3) \text{ m}^3 = 6 \times 6 = 36$ atm m$^3$.
Work = $36 \times 101300 = 3646800$ J.
Area = $(8 - 2) \text{ atm} \times (9 - 3) \text{ m}^3 = 6 \times 6 = 36$ atm m$^3$.
Work = $36 \times 101300 = 3646800$ J.
Q59.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 3 m$^3$ and 9 m$^3$, and the pressures alternate between 2 atm and 6 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(6 - 2) \text{ atm} \times (9 - 3) \text{ m}^3 = 4 \times 6 = 24$ atm m$^3$.
Work = $24 \times 101300 = 2431200$ J.
Area = $(6 - 2) \text{ atm} \times (9 - 3) \text{ m}^3 = 4 \times 6 = 24$ atm m$^3$.
Work = $24 \times 101300 = 2431200$ J.
Q60.
In a cyclic process plotted on a P-V diagram, the cycle forms a rectangle. The volumes alternate between 4 m$^3$ and 12 m$^3$, and the pressures alternate between 3 atm and 12 atm. If the cycle is clockwise, find the net work done.
Area of rectangle = $\Delta P \times \Delta V$.
Area = $(12 - 3) \text{ atm} \times (12 - 4) \text{ m}^3 = 9 \times 8 = 72$ atm m$^3$.
Work = $72 \times 101300 = 7293600$ J.
Area = $(12 - 3) \text{ atm} \times (12 - 4) \text{ m}^3 = 9 \times 8 = 72$ atm m$^3$.
Work = $72 \times 101300 = 7293600$ J.
Topic 7: Heat Engines
Q61.
A heat engine absorbs 1841 J of heat from a hot reservoir and produces 395 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 395 / 1841 = 21.46\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1841 - 395 = 1446$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1841 - 395 = 1446$ J.
Q62.
A heat engine absorbs 1537 J of heat from a hot reservoir and produces 355 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 355 / 1537 = 23.10\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1537 - 355 = 1182$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1537 - 355 = 1182$ J.
Q63.
A heat engine absorbs 1533 J of heat from a hot reservoir and produces 254 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 254 / 1533 = 16.57\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1533 - 254 = 1279$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1533 - 254 = 1279$ J.
Q64.
A heat engine absorbs 1911 J of heat from a hot reservoir and produces 310 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 310 / 1911 = 16.22\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1911 - 310 = 1601$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1911 - 310 = 1601$ J.
Q65.
A heat engine absorbs 1016 J of heat from a hot reservoir and produces 212 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 212 / 1016 = 20.87\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1016 - 212 = 804$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1016 - 212 = 804$ J.
Q66.
A heat engine absorbs 691 J of heat from a hot reservoir and produces 180 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 180 / 691 = 26.05\%$.
Heat rejected $Q_{out} = Q_{in} - W = 691 - 180 = 511$ J.
Heat rejected $Q_{out} = Q_{in} - W = 691 - 180 = 511$ J.
Q67.
A heat engine absorbs 1766 J of heat from a hot reservoir and produces 208 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 208 / 1766 = 11.78\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1766 - 208 = 1558$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1766 - 208 = 1558$ J.
Q68.
A heat engine absorbs 1563 J of heat from a hot reservoir and produces 153 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 153 / 1563 = 9.79\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1563 - 153 = 1410$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1563 - 153 = 1410$ J.
Q69.
A heat engine absorbs 801 J of heat from a hot reservoir and produces 378 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 378 / 801 = 47.19\%$.
Heat rejected $Q_{out} = Q_{in} - W = 801 - 378 = 423$ J.
Heat rejected $Q_{out} = Q_{in} - W = 801 - 378 = 423$ J.
Q70.
A heat engine absorbs 1119 J of heat from a hot reservoir and produces 157 J of useful work. Calculate its efficiency and the heat rejected to the sink.
Efficiency $\eta = W / Q_{in} = 157 / 1119 = 14.03\%$.
Heat rejected $Q_{out} = Q_{in} - W = 1119 - 157 = 962$ J.
Heat rejected $Q_{out} = Q_{in} - W = 1119 - 157 = 962$ J.
Topic 8: Refrigerators and Heat Pumps
Q71.
A refrigerator's compressor does 133 J of work per cycle to extract 574 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 574 / 133 = 4.32$.
Heat rejected $Q_1 = Q_2 + W = 574 + 133 = 707$ J.
Heat rejected $Q_1 = Q_2 + W = 574 + 133 = 707$ J.
Q72.
A refrigerator's compressor does 123 J of work per cycle to extract 675 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 675 / 123 = 5.49$.
Heat rejected $Q_1 = Q_2 + W = 675 + 123 = 798$ J.
Heat rejected $Q_1 = Q_2 + W = 675 + 123 = 798$ J.
Q73.
A refrigerator's compressor does 192 J of work per cycle to extract 345 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 345 / 192 = 1.80$.
Heat rejected $Q_1 = Q_2 + W = 345 + 192 = 537$ J.
Heat rejected $Q_1 = Q_2 + W = 345 + 192 = 537$ J.
Q74.
A refrigerator's compressor does 61 J of work per cycle to extract 799 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 799 / 61 = 13.10$.
Heat rejected $Q_1 = Q_2 + W = 799 + 61 = 860$ J.
Heat rejected $Q_1 = Q_2 + W = 799 + 61 = 860$ J.
Q75.
A refrigerator's compressor does 230 J of work per cycle to extract 272 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 272 / 230 = 1.18$.
Heat rejected $Q_1 = Q_2 + W = 272 + 230 = 502$ J.
Heat rejected $Q_1 = Q_2 + W = 272 + 230 = 502$ J.
Q76.
A refrigerator's compressor does 88 J of work per cycle to extract 353 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 353 / 88 = 4.01$.
Heat rejected $Q_1 = Q_2 + W = 353 + 88 = 441$ J.
Heat rejected $Q_1 = Q_2 + W = 353 + 88 = 441$ J.
Q77.
A refrigerator's compressor does 245 J of work per cycle to extract 445 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 445 / 245 = 1.82$.
Heat rejected $Q_1 = Q_2 + W = 445 + 245 = 690$ J.
Heat rejected $Q_1 = Q_2 + W = 445 + 245 = 690$ J.
Q78.
A refrigerator's compressor does 75 J of work per cycle to extract 608 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 608 / 75 = 8.11$.
Heat rejected $Q_1 = Q_2 + W = 608 + 75 = 683$ J.
Heat rejected $Q_1 = Q_2 + W = 608 + 75 = 683$ J.
Q79.
A refrigerator's compressor does 181 J of work per cycle to extract 638 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 638 / 181 = 3.52$.
Heat rejected $Q_1 = Q_2 + W = 638 + 181 = 819$ J.
Heat rejected $Q_1 = Q_2 + W = 638 + 181 = 819$ J.
Q80.
A refrigerator's compressor does 111 J of work per cycle to extract 490 J of heat from the cold interior. Calculate its Coefficient of Performance (COP) and the heat rejected to the room.
COP $\alpha = Q_2 / W = 490 / 111 = 4.41$.
Heat rejected $Q_1 = Q_2 + W = 490 + 111 = 601$ J.
Heat rejected $Q_1 = Q_2 + W = 490 + 111 = 601$ J.
Topic 9: Carnot Engine Efficiency
Q81.
A Carnot engine operates between a hot reservoir at 162$^\circ$C and a cold sink at 26$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 435$ K, $T_2 = 299$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 299/435 = 31.26\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 299/435 = 31.26\%$.
Q82.
A Carnot engine operates between a hot reservoir at 194$^\circ$C and a cold sink at 28$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 467$ K, $T_2 = 301$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 301/467 = 35.55\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 301/467 = 35.55\%$.
Q83.
A Carnot engine operates between a hot reservoir at 156$^\circ$C and a cold sink at 40$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 429$ K, $T_2 = 313$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 313/429 = 27.04\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 313/429 = 27.04\%$.
Q84.
A Carnot engine operates between a hot reservoir at 157$^\circ$C and a cold sink at 37$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 430$ K, $T_2 = 310$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 310/430 = 27.91\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 310/430 = 27.91\%$.
Q85.
A Carnot engine operates between a hot reservoir at 219$^\circ$C and a cold sink at 35$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 492$ K, $T_2 = 308$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 308/492 = 37.40\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 308/492 = 37.40\%$.
Q86.
A Carnot engine operates between a hot reservoir at 140$^\circ$C and a cold sink at 40$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 413$ K, $T_2 = 313$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 313/413 = 24.21\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 313/413 = 24.21\%$.
Q87.
A Carnot engine operates between a hot reservoir at 229$^\circ$C and a cold sink at 40$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 502$ K, $T_2 = 313$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 313/502 = 37.65\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 313/502 = 37.65\%$.
Q88.
A Carnot engine operates between a hot reservoir at 271$^\circ$C and a cold sink at 44$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 544$ K, $T_2 = 317$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 317/544 = 41.73\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 317/544 = 41.73\%$.
Q89.
A Carnot engine operates between a hot reservoir at 102$^\circ$C and a cold sink at 25$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 375$ K, $T_2 = 298$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 298/375 = 20.53\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 298/375 = 20.53\%$.
Q90.
A Carnot engine operates between a hot reservoir at 208$^\circ$C and a cold sink at 22$^\circ$C. Calculate its theoretical maximum efficiency.
Convert to Kelvin: $T_1 = 481$ K, $T_2 = 295$ K.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 295/481 = 38.67\%$.
Carnot Efficiency $\eta = 1 - T_2/T_1 = 1 - 295/481 = 38.67\%$.
Topic 10: Second Law Concepts & Entropy
Q91.
An isothermal reservoir at 300 K supplies 2512 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -2512$ J.
Entropy change $\Delta S = \Delta Q / T = -2512 / 300 = -8.37$ J/K.
Entropy change $\Delta S = \Delta Q / T = -2512 / 300 = -8.37$ J/K.
Q92.
An isothermal reservoir at 273 K supplies 4292 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -4292$ J.
Entropy change $\Delta S = \Delta Q / T = -4292 / 273 = -15.72$ J/K.
Entropy change $\Delta S = \Delta Q / T = -4292 / 273 = -15.72$ J/K.
Q93.
An isothermal reservoir at 300 K supplies 3059 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -3059$ J.
Entropy change $\Delta S = \Delta Q / T = -3059 / 300 = -10.20$ J/K.
Entropy change $\Delta S = \Delta Q / T = -3059 / 300 = -10.20$ J/K.
Q94.
An isothermal reservoir at 273 K supplies 2506 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -2506$ J.
Entropy change $\Delta S = \Delta Q / T = -2506 / 273 = -9.18$ J/K.
Entropy change $\Delta S = \Delta Q / T = -2506 / 273 = -9.18$ J/K.
Q95.
An isothermal reservoir at 273 K supplies 1532 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -1532$ J.
Entropy change $\Delta S = \Delta Q / T = -1532 / 273 = -5.61$ J/K.
Entropy change $\Delta S = \Delta Q / T = -1532 / 273 = -5.61$ J/K.
Q96.
An isothermal reservoir at 373 K supplies 3203 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -3203$ J.
Entropy change $\Delta S = \Delta Q / T = -3203 / 373 = -8.59$ J/K.
Entropy change $\Delta S = \Delta Q / T = -3203 / 373 = -8.59$ J/K.
Q97.
An isothermal reservoir at 300 K supplies 1343 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -1343$ J.
Entropy change $\Delta S = \Delta Q / T = -1343 / 300 = -4.48$ J/K.
Entropy change $\Delta S = \Delta Q / T = -1343 / 300 = -4.48$ J/K.
Q98.
An isothermal reservoir at 373 K supplies 4049 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -4049$ J.
Entropy change $\Delta S = \Delta Q / T = -4049 / 373 = -10.86$ J/K.
Entropy change $\Delta S = \Delta Q / T = -4049 / 373 = -10.86$ J/K.
Q99.
An isothermal reservoir at 300 K supplies 2638 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -2638$ J.
Entropy change $\Delta S = \Delta Q / T = -2638 / 300 = -8.79$ J/K.
Entropy change $\Delta S = \Delta Q / T = -2638 / 300 = -8.79$ J/K.
Q100.
An isothermal reservoir at 273 K supplies 3852 J of heat to a reversible engine. What is the change in entropy of the reservoir?
The reservoir loses heat, so $\Delta Q = -3852$ J.
Entropy change $\Delta S = \Delta Q / T = -3852 / 273 = -14.11$ J/K.
Entropy change $\Delta S = \Delta Q / T = -3852 / 273 = -14.11$ J/K.