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  4. Thermodynamics Chapter-Wise Test 2

Thermodynamics Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A gas at \( 11 \, \text{atm} \) and \( 90^\circ \text{C} \) in a \( 10 \, \text{L} \) container is cooled isochorically to \( 30^\circ \text{C} \). What is the final pressure?

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

\( P_1 = 11 \, \text{atm} \), \( T_1 = 90 + 273 = 363 \, \text{K} \), \( T_2 = 30 + 273 = 303 \, \text{K} \).

\( \frac{11}{363} = \frac{P_2}{303} \Rightarrow P_2 = \frac{11 \times 303}{363} \approx 9.18 \, \text{atm} \).

8.5 atm
9.2 atm
9.5 atm
10.0 atm
2

A gas undergoes an adiabatic expansion from \( 28 \, \text{L} \) to \( 84 \, \text{L} \), reducing its pressure from \( 15 \, \text{atm} \) to \( 3 \, \text{atm} \). What is the value of \( \gamma \)?

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

\( 15 \times 28^\gamma = 3 \times 84^\gamma \).

\( 15 / 3 = \left(\frac{84}{28}\right)^\gamma \Rightarrow 5 = 3^\gamma \).

\( \gamma = \frac{\log(5)}{\log(3)} \approx 1.465 \approx 1.5 \) (standard value from context).

1.33
1.5
1.67
2.0
2

How many calories are equivalent to \( 1674 \, \text{J} \) of heat? (1 cal = \( 4.186 \, \text{J} \))

\( \text{Heat in cal} = \frac{\text{Heat in J}}{4.186} \).

\( \frac{1674}{4.186} \approx 399.9 \approx 400 \, \text{cal} \).

350 cal
400 cal
450 cal
500 cal
2

A system absorbs \( 730 \, \text{J} \) of heat and performs \( 190 \, \text{J} \) of work. What is the change in internal energy?

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

\( \Delta Q = 730 \), \( \Delta W = 190 \) (work by system).

\( 730 = \Delta U + 190 \Rightarrow \Delta U = 730 - 190 = 540 \, \text{J} \).

500 J
540 J
600 J
650 J
2

In an isobaric process, \( 1.1 \, \text{moles} \) of an ideal gas expand from \( 7 \, \text{L} \) to \( 14 \, \text{L} \) at \( 390 \, \text{K} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \Delta V = 14 - 7 = 7 \, \text{L} \).

\( P = \frac{\mu R T}{V_1} = \frac{1.1 \times 8.3 \times 390}{7} = 509.01 \, \text{atm} \) (unit correction needed).

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

\( W = 1.1 \times 8.3 \times 390 = 3560.7 \, \text{J} \approx 3561 \, \text{J} \) (adjusted).

3400 J
3561 J
3700 J
3900 J
2

In an isobaric process, \( 0.5 \, \text{moles} \) of gas expand from \( 300 \, \text{K} \) to \( 450 \, \text{K} \). What is the heat supplied if \( C_p = 29.1 \, \text{J mol}^{-1} \text{K}^{-1} \)?

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

\( \mu = 0.5 \), \( C_p = 29.1 \), \( \Delta T = 450 - 300 = 150 \).

\( \Delta Q = 0.5 \times 29.1 \times 150 = 2182.5 \, \text{J} \).

2000 J
2183 J
2300 J
2500 J
2

A monatomic ideal gas undergoes an adiabatic expansion, reducing its temperature from \( 400 \, \text{K} \) to \( 300 \, \text{K} \). How much work is done by the gas if \( \mu = 1 \, \text{mole} \) and \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \)? (\( \gamma = 1.67 \))

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

\( \mu = 1 \), \( R = 8.3 \), \( T_1 = 400 \), \( T_2 = 300 \), \( \gamma = 1.67 \).

\( W = \frac{1 \times 8.3 \times (400 - 300)}{1.67 - 1} = \frac{8.3 \times 100}{0.67} \approx 1238.8 \, \text{J} \).

1000 J
1239 J
1500 J
2000 J
2

Which of the following statements is correct about a reversible process?

A reversible process can be reversed, restoring the system and surroundings to their original states without net external changes, requiring quasi-static conditions and no dissipation. Option D is correct.

It involves rapid changes
It always increases internal energy
It requires friction
It restores original states of system and surroundings
4

A gas at \( 10 \, \text{atm} \) and \( 80^\circ \text{C} \) in a \( 8 \, \text{L} \) container is heated isochorically to \( 140^\circ \text{C} \). What is the final pressure?

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

\( P_1 = 10 \, \text{atm} \), \( T_1 = 80 + 273 = 353 \, \text{K} \), \( T_2 = 140 + 273 = 413 \, \text{K} \).

\( \frac{10}{353} = \frac{P_2}{413} \Rightarrow P_2 = \frac{10 \times 413}{353} \approx 11.7 \, \text{atm} \).

11.0 atm
11.7 atm
12.0 atm
12.5 atm
2

In an adiabatic process, what compensates for the absence of heat transfer?

In an adiabatic process (\( \Delta Q = 0 \)), the change in internal energy (\( \Delta U \)) is entirely due to work done (\( \Delta U = -\Delta W \)). Work done by or on the system adjusts the internal energy, as no heat is exchanged.

Temperature increase
Work done
Pressure change
Volume expansion
2

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