Thermodynamics Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following statements is incorrect about the Zeroth Law of Thermodynamics?

The Zeroth Law defines temperature through thermal equilibrium but does not address heat flow direction (Second Law) or energy conservation (First Law). Option C is incorrect.

It establishes temperature as a measurable property
It applies to systems in thermal equilibrium
It governs the direction of heat flow
It involves a third system for comparison
3

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

In an isothermal process (\( T = \text{constant} \)), the internal energy of an ideal gas (\( U \), temperature-dependent) remains constant (\( \Delta U = 0 \)), and heat supplied equals work done (\( \Delta Q = \Delta W \)), making option C correct.

Internal energy increases
No heat is exchanged
Heat supplied equals work done
Pressure remains constant
3

What is the molar specific heat capacity at constant pressure for a diatomic gas if \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \)?

For diatomic gas: \( C_v = \frac{5}{2} R \), \( C_p = C_v + R = \frac{7}{2} R \).

\( C_p = \frac{7}{2} \times 8.3 = 29.05 \, \text{J mol}^{-1} \text{K}^{-1} \approx 29.1 \, \text{J mol}^{-1} \text{K}^{-1} \).

20.8 J mol⁻¹ K⁻¹
25.0 J mol⁻¹ K⁻¹
29.1 J mol⁻¹ K⁻¹
33.2 J mol⁻¹ K⁻¹
3

0.1 kg of a substance absorbs 1200 J of heat, increasing its temperature from 20°C to 50°C. What is its specific heat capacity?

Specific heat: \( s = \frac{\Delta Q}{m \Delta T} \). Given \( \Delta Q = 1200 \, \text{J} \), \( m = 0.1 \, \text{kg} \), \( \Delta T = 50 - 20 = 30 \, \text{K} \). \( s = \frac{1200}{0.1 \times 30} = 400 \, \text{J kg}^{-1} \text{K}^{-1} \).

300 J kg⁻¹ K⁻¹
400 J kg⁻¹ K⁻¹
500 J kg⁻¹ K⁻¹
600 J kg⁻¹ K⁻¹
2

How many joules are equivalent to \( 250 \, \text{cal} \) of heat? (1 cal = \( 4.186 \, \text{J} \))

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

\( 250 \times 4.186 = 1046.5 \, \text{J} \approx 1047 \, \text{J} \).

900 J
1047 J
1100 J
1200 J
2

How much heat is required to vaporize \( 0.5 \, \text{g} \) of water at \( 100^\circ \text{C} \) and \( 1 \, \text{atm} \)? (Latent heat = \( 2256 \, \text{J/g} \))

\( \Delta Q = m L \).

\( m = 0.5 \), \( L = 2256 \).

\( \Delta Q = 0.5 \times 2256 = 1128 \, \text{J} \).

1000 J
1128 J
1200 J
1300 J
2

How much heat is required to raise the temperature of \( 0.15 \, \text{kg} \) of copper from \( 40^\circ \text{C} \) to \( 60^\circ \text{C} \)? (Specific heat of copper = \( 386.4 \, \text{J kg}^{-1} \text{K}^{-1} \))

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

\( m = 0.15 \), \( s = 386.4 \), \( \Delta T = 60 - 40 = 20 \).

\( \Delta Q = 0.15 \times 386.4 \times 20 = 1159.2 \, \text{J} \approx 1159 \, \text{J} \).

1000 J
1159 J
1200 J
1300 J
2

A diatomic gas undergoes an adiabatic expansion from \( 780 \, \text{K} \) to \( 390 \, \text{K} \) with \( 1.3 \, \text{moles} \). What is the work done? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \), \( \gamma = 1.4 \))

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

\( \mu = 1.3 \), \( R = 8.3 \), \( T_1 = 780 \), \( T_2 = 390 \), \( \gamma = 1.4 \).

\( W = \frac{1.3 \times 8.3 \times (780 - 390)}{1.4 - 1} = \frac{10.79 \times 390}{0.4} = 10520.25 \, \text{J} \approx 10520 \, \text{J} \).

10000 J
10520 J
11000 J
11500 J
2

A system releases \( 720 \, \text{J} \) of heat and has \( 180 \, \text{J} \) of work done on it. What is the change in internal energy?

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

\( \Delta Q = -720 \) (heat released), \( \Delta W = -180 \) (work on system).

\( -720 = \Delta U - 180 \Rightarrow \Delta U = -720 + 180 = -540 \, \text{J} \).

-600 J
-540 J
-500 J
-450 J
2

In an isobaric process, \( 1.4 \, \text{moles} \) of an ideal gas expand from \( 5 \, \text{L} \) to \( 11 \, \text{L} \) at \( 370 \, \text{K} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \Delta V = 11 - 5 = 6 \, \text{L} \).

\( P = \frac{\mu R T}{V_1} = \frac{1.4 \times 8.3 \times 370}{5} = 860.44 \, \text{atm} \) (unit correction needed).

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

\( W = 1.4 \times 8.3 \times 370 = 4302.2 \, \text{J} \approx 4302 \, \text{J} \) (adjusted).

4000 J
4302 J
4500 J
4700 J
2

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