Thermodynamics Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following statements is correct about the Second Law of Thermodynamics?

The Second Law (Clausius statement) states that heat cannot flow from a colder to a hotter body without work, reflecting natural directionality. Option C is correct.

Heat flows spontaneously from cold to hot
All processes are reversible
Work is needed to transfer heat from cold to hot
Internal energy remains constant
3

Which of the following statements is correct about isothermal processes for an ideal gas?

In an isothermal process (\( T = \text{constant} \)), \( \Delta U = 0 \) for an ideal gas, and \( P V = \text{constant} \). Option C is correct.

Internal energy decreases
Pressure remains constant
\( P V \) remains constant
No work is done
3

A gas at \( 6 \, \text{atm} \) and \( 50^\circ \text{C} \) in a \( 4 \, \text{L} \) container is heated isochorically to \( 150^\circ \text{C} \). What is the final pressure?

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

\( P_1 = 6 \, \text{atm} \), \( T_1 = 50 + 273 = 323 \, \text{K} \), \( T_2 = 150 + 273 = 423 \, \text{K} \).

\( \frac{6}{323} = \frac{P_2}{423} \Rightarrow P_2 = \frac{6 \times 423}{323} \approx 7.86 \, \text{atm} \).

7.0 atm
7.9 atm
8.5 atm
9.0 atm
2

How much heat is required to raise the temperature of \( 0.2 \, \text{kg} \) of silver from \( 30^\circ \text{C} \) to \( 50^\circ \text{C} \)? (Specific heat of silver = \( 236.1 \, \text{J kg}^{-1} \text{K}^{-1} \))

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

\( m = 0.2 \), \( s = 236.1 \), \( \Delta T = 50 - 30 = 20 \).

\( \Delta Q = 0.2 \times 236.1 \times 20 = 944.4 \, \text{J} \approx 944 \, \text{J} \).

800 J
944 J
1000 J
1200 J
2

A system absorbs \( 710 \, \text{J} \) of heat and performs \( 260 \, \text{J} \) of work. What is the change in internal energy?

First Law: \( \Delta Q = \Delta U + \Delta W \).

\( \Delta Q = 710 \), \( \Delta W = 260 \) (work by system).

\( 710 = \Delta U + 260 \Rightarrow \Delta U = 710 - 260 = 450 \, \text{J} \).

400 J
450 J
500 J
550 J
2

How much work is done by \( 1 \, \text{mole} \) of an ideal gas in an isothermal expansion from \( 3 \, \text{L} \) to \( 9 \, \text{L} \) at \( 500 \, \text{K} \)? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 1 \), \( R = 8.3 \), \( T = 500 \), \( V_2 = 9 \), \( V_1 = 3 \).

\( W = 1 \times 8.3 \times 500 \times \ln\left(\frac{9}{3}\right) = 4150 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 4150 \times 1.0986 \approx 4559 \, \text{J} \).

4000 J
4560 J
5000 J
5500 J
2

Why does an isobaric process involve both internal energy change and work?

In an isobaric process (\( P = \text{constant} \)), heat added (\( \Delta Q \)) increases internal energy (\( \Delta U \)) and does work (\( W = P \Delta V \)) due to volume expansion, as per the First Law: \( \Delta Q = \Delta U + P \Delta V \).

Volume remains constant
Heat supplies both energy and work
Temperature decreases
No heat is added
2

What is the significance of a quasi-static process in thermodynamics?

A quasi-static process is infinitely slow, ensuring the system remains in thermal and mechanical equilibrium with its surroundings at every stage. This allows well-defined state variables (e.g., \( P \), \( T \)) and is an idealized condition for reversible processes.

It ensures rapid changes
It maintains equilibrium at every stage
It prevents work from being done
It eliminates heat transfer
2

A gas undergoes an adiabatic compression from \( 20 \, \text{L} \) to \( 5 \, \text{L} \), increasing its pressure from \( 2 \, \text{atm} \) to \( 16 \, \text{atm} \). What is the value of \( \gamma \)?

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

\( 2 \times 20^\gamma = 16 \times 5^\gamma \).

\( \frac{20^\gamma}{5^\gamma} = \frac{16}{2} \Rightarrow \left(\frac{20}{5}\right)^\gamma = 8 \Rightarrow 4^\gamma = 8 \).

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

1.33
1.5
1.67
2.0
2

A system releases \( 300 \, \text{J} \) of heat and has \( 150 \, \text{J} \) of work done on it. What is the change in internal energy?

First Law: \( \Delta Q = \Delta U + \Delta W \).

\( \Delta Q = -300 \, \text{J} \) (heat released), \( \Delta W = -150 \, \text{J} \) (work done on system, negative by convention).

\( -300 = \Delta U + (-150) \).

\( \Delta U = -300 + 150 = -150 \, \text{J} \).

-150 J
-100 J
100 J
150 J
1

A gas at \( 7 \, \text{atm} \) and \( 40^\circ \text{C} \) in a \( 5 \, \text{L} \) container is cooled isochorically to \( -20^\circ \text{C} \). What is the final pressure?

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

\( P_1 = 7 \, \text{atm} \), \( T_1 = 40 + 273 = 313 \, \text{K} \), \( T_2 = -20 + 273 = 253 \, \text{K} \).

\( \frac{7}{313} = \frac{P_2}{253} \Rightarrow P_2 = \frac{7 \times 253}{313} \approx 5.66 \, \text{atm} \).

5.0 atm
5.7 atm
6.0 atm
6.5 atm
2

Which of the following statements is incorrect about an isochoric process?

In an isochoric process (\( \Delta V = 0 \)), no work is done (\( W = 0 \)), and heat changes internal energy (\( \Delta Q = \Delta U \)). Option B is incorrect; work is not done by the gas.

Volume remains constant
Work is done by the gas
Heat changes internal energy
Temperature can change
2

In an isobaric process, \( 0.7 \, \text{moles} \) of gas expand from \( 320 \, \text{K} \) to \( 400 \, \text{K} \). What is the heat supplied if \( C_p = 29.1 \, \text{J mol}^{-1} \text{K}^{-1} \)?

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

\( \mu = 0.7 \), \( C_p = 29.1 \), \( \Delta T = 400 - 320 = 80 \).

\( \Delta Q = 0.7 \times 29.1 \times 80 = 1632 \, \text{J} \).

1500 J
1632 J
1700 J
1800 J
2

How much heat is required to raise the temperature of \( 0.45 \, \text{kg} \) of aluminium from \( 30^\circ \text{C} \) to \( 60^\circ \text{C} \)? (Specific heat of aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \))

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

\( m = 0.45 \), \( s = 900 \), \( \Delta T = 60 - 30 = 30 \).

\( \Delta Q = 0.45 \times 900 \times 30 = 12150 \, \text{J} \).

11500 J
12150 J
12500 J
13000 J
2

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