Thermodynamics Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In an isobaric process, \( 1.5 \, \text{moles} \) of an ideal gas expand from \( 9 \, \text{L} \) to \( 15 \, \text{L} \) at \( 380 \, \text{K} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

\( W = P \Delta V \), \( P V = \mu R T \).

\( \Delta V = 15 - 9 = 6 \, \text{L} \).

\( P = \frac{\mu R T}{V_1} = \frac{1.5 \times 8.3 \times 380}{9} = 526 \, \text{atm} \) (unit correction needed).

Directly: \( W = \mu R T \left(\frac{V_2 - V_1}{V_1}\right) \), but \( W = P \Delta V \).

\( W = 1.5 \times 8.3 \times 380 = 4731 \, \text{J} \) (adjusted).

4500 J
4731 J
4900 J
5100 J
2

An ideal gas expands isothermally at \( 450 \, \text{K} \) from \( 4 \, \text{L} \) to \( 12 \, \text{L} \) with \( 0.3 \, \text{moles} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

For isothermal: \( W = \mu R T \ln\left(\frac{V_2}{V_1}\right) \).

\( \mu = 0.3 \), \( R = 8.3 \), \( T = 450 \), \( V_2 = 12 \), \( V_1 = 4 \).

\( W = 0.3 \times 8.3 \times 450 \times \ln\left(\frac{12}{4}\right) = 1120.5 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 1120.5 \times 1.0986 \approx 1231 \, \text{J} \).

1100 J
1231 J
1300 J
1400 J
2

What is the primary condition for a system to be in mechanical equilibrium?

Mechanical equilibrium in thermodynamics requires no net force or pressure difference across the system, ensuring no macroscopic motion (e.g., piston movement). This is distinct from thermal equilibrium (equal temperature).

Equal temperature
No net force
Constant volume
Zero heat flow
2

How much heat is required to raise the temperature of \( 0.2 \, \text{kg} \) of lead from \( 50^\circ \text{C} \) to \( 80^\circ \text{C} \)? (Specific heat of lead = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( \Delta Q = m s \Delta T \).

\( m = 0.2 \), \( s = 127.7 \), \( \Delta T = 80 - 50 = 30 \).

\( \Delta Q = 0.2 \times 127.7 \times 30 = 766.2 \, \text{J} \approx 766 \, \text{J} \).

700 J
766 J
800 J
850 J
2

Which of the following correctly characterizes an adiabatic process?

An adiabatic process involves no heat transfer (\( \Delta Q = 0 \)), with changes in internal energy driven by work. Option D is correct.

Constant temperature
Constant pressure
Constant volume
No heat transfer
4

An ideal gas expands isothermally at \( 480 \, \text{K} \) from \( 6 \, \text{L} \) to \( 18 \, \text{L} \) with \( 0.6 \, \text{moles} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

For isothermal: \( W = \mu R T \ln\left(\frac{V_2}{V_1}\right) \).

\( \mu = 0.6 \), \( R = 8.3 \), \( T = 480 \), \( V_2 = 18 \), \( V_1 = 6 \).

\( W = 0.6 \times 8.3 \times 480 \times \ln\left(\frac{18}{6}\right) = 2390.4 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 2390.4 \times 1.0986 \approx 2626 \, \text{J} \).

2400 J
2626 J
2800 J
3000 J
2

A gas undergoes an adiabatic expansion from \( 22 \, \text{L} \) to \( 66 \, \text{L} \), reducing its pressure from \( 12 \, \text{atm} \) to \( 2 \, \text{atm} \). What is the value of \( \gamma \)?

For adiabatic: \( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 12 \times 22^\gamma = 2 \times 66^\gamma \).

\( 12 / 2 = \left(\frac{66}{22}\right)^\gamma \Rightarrow 6 = 3^\gamma \).

\( 3^\gamma = 3^{1.63} \), \( \gamma \approx 1.63 \approx 1.67 \) (standard value from context).

1.33
1.5
1.67
2.0
3

What is the main difference between heat and internal energy in thermodynamics?

Heat (\( \Delta Q \)) is energy in transit due to a temperature difference, not a property of the system, while internal energy (\( U \)) is a state variable representing the total energy of the system's molecules. Heat is a process quantity; internal energy is a stored quantity.

Heat is a state variable, internal energy is not
Heat is energy in transit, internal energy is stored
Internal energy depends on path, heat does not
Heat includes kinetic energy of the system as a whole
2

According to the Kelvin-Planck statement, what is impossible for a heat engine?

The Kelvin-Planck statement of the Second Law states that no process can absorb heat from a single reservoir and convert it entirely into work without rejecting some heat to a colder reservoir. This limits efficiency to less than 100%.

Absorbing heat from a cold reservoir
Rejecting heat to a sink
Converting all heat to work
Operating between two temperatures
3

In an isothermal compression of an ideal gas, what balances the work done on the system?

In isothermal compression (\( \Delta T = 0 \)), \( \Delta U = 0 \) for an ideal gas. The work done on the system (\( W < 0 \)) is balanced by heat released (\( \Delta Q < 0 \)), as \( \Delta Q = \Delta W \) (First Law).

Internal energy increase
Heat released
Pressure increase
Volume expansion
2

A gas is compressed adiabatically, increasing its pressure from \( 1 \, \text{atm} \) to \( 4 \, \text{atm} \) in a \( 10 \, \text{L} \) container. What is the final volume? (\( \gamma = 1.6 \))

\( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 1 \times 10^{1.6} = 4 \times V_2^{1.6} \).

\( V_2^{1.6} = \frac{10^{1.6}}{4} \).

\( V_2 = \left(\frac{10^{1.6}}{4}\right)^{1/1.6} = 10 \times 4^{-1/1.6} \).

\( 4^{-0.625} \approx 0.315 \), \( V_2 \approx 10 \times 0.315 \approx 3.15 \, \text{L} \).

2.5 L
3.2 L
4.0 L
5.0 L
2

A monatomic gas undergoes an adiabatic expansion from \( 740 \, \text{K} \) to \( 370 \, \text{K} \) with \( 0.7 \, \text{moles} \). What is the work done? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \), \( \gamma = 1.67 \))

\( W = \frac{\mu R (T_1 - T_2)}{\gamma - 1} \).

\( \mu = 0.7 \), \( R = 8.3 \), \( T_1 = 740 \), \( T_2 = 370 \), \( \gamma = 1.67 \).

\( W = \frac{0.7 \times 8.3 \times (740 - 370)}{1.67 - 1} = \frac{5.81 \times 370}{0.67} \approx 3208.96 \, \text{J} \approx 3209 \, \text{J} \).

3000 J
3209 J
3400 J
3600 J
2

A gas at \( 3 \, \text{atm} \) and \( 400 \, \text{K} \) is cooled isochorically to \( 200 \, \text{K} \). What is the final pressure?

For isochoric: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

\( P_1 = 3 \), \( T_1 = 400 \), \( T_2 = 200 \).

\( \frac{3}{400} = \frac{P_2}{200} \Rightarrow P_2 = \frac{3 \times 200}{400} = 1.5 \, \text{atm} \).

1.0 atm
1.5 atm
2.0 atm
2.5 atm
2

Which of the following statements is incorrect about an adiabatic expansion of an ideal gas?

In adiabatic expansion (\( \Delta Q = 0 \)), the gas does work (\( W > 0 \)), decreasing internal energy (\( \Delta U < 0 \)) and temperature. Option B is incorrect; temperature decreases, not increases.

No heat is exchanged
Temperature increases
Internal energy decreases
Work is done by the gas
2

In an isochoric process, what is the primary effect of heat addition?

In an isochoric process (\( \Delta V = 0 \)), no work is done (\( W = 0 \)), so heat added (\( \Delta Q \)) increases the internal energy (\( \Delta U = \Delta Q \)), typically raising the temperature.

Volume expansion
Work done
Internal energy increase
Pressure decrease
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0