Thermodynamics Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In an isobaric process, \( 1.6 \, \text{moles} \) of an ideal gas expand from \( 6 \, \text{L} \) to \( 12 \, \text{L} \) at \( 400 \, \text{K} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \Delta V = 12 - 6 = 6 \, \text{L} \).

\( P = \frac{\mu R T}{V_1} = \frac{1.6 \times 8.3 \times 400}{6} = 885.33 \, \text{atm} \) (unit correction needed).

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

\( W = 1.6 \times 8.3 \times 400 = 5312 \, \text{J} \) (adjusted).

5000 J
5312 J
5500 J
5700 J
2

In an isobaric process, what remains constant?

An isobaric process is defined by constant pressure. While volume and temperature may change, pressure remains fixed throughout the process.

Volume
Temperature
Pressure
Internal energy
3

Why does the Second Law impose a limit on the efficiency of a heat engine?

The Second Law (e.g., Kelvin-Planck) requires some heat to be rejected to a cold reservoir, preventing complete conversion of heat to work. This inherent loss sets a maximum efficiency below 100%, dependent on temperature difference.

Energy is conserved
Heat must be rejected
Work equals heat
Temperature remains constant
2

Which of the following statements correctly defines thermal equilibrium?

Thermal equilibrium occurs when two systems in contact have no net heat flow between them, indicating equal temperatures, as per the Zeroth Law. Option C is correct.

Systems have different temperatures
Heat flows continuously between systems
No net heat flows due to equal temperatures
Pressure remains unequal
3

A gas at \( 2 \, \text{atm} \) and \( 27^\circ \text{C} \) in a \( 5 \, \text{L} \) container is compressed isochorically until its pressure becomes \( 4 \, \text{atm} \). What is the final temperature?

For an isochoric process, \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

\( P_1 = 2 \, \text{atm} \), \( T_1 = 27 + 273 = 300 \, \text{K} \), \( P_2 = 4 \, \text{atm} \).

\( \frac{2}{300} = \frac{4}{T_2} \Rightarrow T_2 = \frac{4 \times 300}{2} = 600 \, \text{K} \).

\( T_2 = 600 - 273 = 327^\circ \text{C} \).

300°C
327°C
350°C
373°C
2

Which of the following is NOT a requirement for the Zeroth Law to apply?

The Zeroth Law requires thermal equilibrium (no heat flow, equal temperatures) between systems, not equal pressures or work exchange. Option B is not a requirement.

No net heat flow
Equal pressures
Equal temperatures
Contact between systems
2

What prevents a spontaneous process like free expansion from being reversible?

Free expansion (e.g., gas into a vacuum) is irreversible because it involves non-equilibrium states with no work done to restore the original state. The system cannot return to its initial condition without external intervention.

Constant temperature
Non-equilibrium states
Heat addition
Pressure equality
2

In an isobaric process, \( 1.2 \, \text{moles} \) of gas expand from \( 340 \, \text{K} \) to \( 430 \, \text{K} \). What is the heat supplied if \( C_p = 24.9 \, \text{J mol}^{-1} \text{K}^{-1} \)?

\( \Delta Q = \mu C_p \Delta T \).

\( \mu = 1.2 \), \( C_p = 24.9 \), \( \Delta T = 430 - 340 = 90 \).

\( \Delta Q = 1.2 \times 24.9 \times 90 = 2694.6 \, \text{J} \approx 2695 \, \text{J} \).

2500 J
2695 J
2800 J
3000 J
2

Which of the following statements is correct about a quasi-static process?

A quasi-static process is infinitely slow, maintaining the system in equilibrium with its surroundings at every stage, allowing well-defined \( P \) and \( T \). Option A is correct; it’s a condition for reversibility but not sufficient alone.

It occurs in equilibrium at every stage
It always involves heat transfer
It is inherently irreversible
It requires rapid changes
1

Which of the following statements is correct about heat in thermodynamics?

Heat (\( \Delta Q \)) is energy transferred due to a temperature difference, not a stored quantity or state variable. It is path-dependent, unlike internal energy, making option B correct.

It is a state variable
It is energy transferred due to temperature difference
It is always equal to work done
It depends only on the initial state
2

What is the change in internal energy for \( 0.7 \, \text{moles} \) of an ideal gas heated from \( 280 \, \text{K} \) to \( 340 \, \text{K} \) at constant volume? (\( C_v = 20.8 \, \text{J mol}^{-1} \text{K}^{-1} \))

\( \Delta U = \mu C_v \Delta T \).

\( \mu = 0.7 \), \( C_v = 20.8 \), \( \Delta T = 340 - 280 = 60 \).

\( \Delta U = 0.7 \times 20.8 \times 60 = 873.6 \, \text{J} \approx 874 \, \text{J} \).

800 J
874 J
900 J
950 J
2

A gas undergoes an adiabatic compression from \( 16 \, \text{L} \) to \( 4 \, \text{L} \), increasing its pressure from \( 1 \, \text{atm} \) to \( 8 \, \text{atm} \). What is the value of \( \gamma \)?

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

\( 1 \times 16^\gamma = 8 \times 4^\gamma \).

\( 16^\gamma = 8 \times 4^\gamma \).

\( \left(\frac{16}{4}\right)^\gamma = 8 \Rightarrow 4^\gamma = 8 \).

\( 4^\gamma = 2^3 \Rightarrow 2^{2\gamma} = 2^3 \Rightarrow 2\gamma = 3 \Rightarrow \gamma = 1.5 \).

1.33
1.5
1.67
2.0
2

In an adiabatic process, \( 1 \, \text{mole} \) of gas at \( 600 \, \text{K} \) does \( 2000 \, \text{J} \) of work. What is the final temperature? (\( \gamma = 1.4 \), \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( 2000 = \frac{1 \times 8.3 \times (600 - T_2)}{1.4 - 1} \).

\( 2000 = \frac{8.3 \times (600 - T_2)}{0.4} \).

\( 2000 \times 0.4 = 8.3 \times (600 - T_2) \Rightarrow 800 = 8.3 \times (600 - T_2) \).

\( 600 - T_2 = \frac{800}{8.3} \approx 96.39 \).

\( T_2 = 600 - 96.39 \approx 503.61 \, \text{K} \approx 504 \, \text{K} \).

480 K
504 K
520 K
550 K
2

A gas expands adiabatically from \( 4 \, \text{atm} \) and \( 8 \, \text{L} \) to \( 1 \, \text{atm} \). What is the final volume? (\( \gamma = 1.67 \))

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

\( 4 \times 8^{1.67} = 1 \times V_2^{1.67} \).

\( V_2^{1.67} = 4 \times 8^{1.67} \).

\( V_2 = (4 \times 8^{1.67})^{1/1.67} = 4^{1/1.67} \times 8 \).

\( 4^{0.5988} \approx 2.3 \), \( 8^{1.67} \approx 29.86 \), but \( V_2 = 8 \times 4^{0.5988} \approx 18.4 \, \text{L} \).

16 L
18.4 L
20 L
22 L
2

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