Thermodynamics Chapter-Wise Test 22

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following is incorrect about reversible processes?

Reversible processes are quasi-static and lack dissipation, not involving rapid changes or friction. Option A is incorrect as rapid changes lead to irreversibility.

They involve rapid changes
They require quasi-static conditions
They have no dissipative effects
They can restore original states
1

Why is the concept of thermal equilibrium essential to the Zeroth Law of Thermodynamics?

The Zeroth Law relies on thermal equilibrium, where no net heat flows between systems in contact, establishing that systems sharing this property with a third system have a common property—temperature—thus defining temperature measurement.

It ensures heat flow
It defines temperature equality
It measures work done
It prevents energy conservation
2

How much heat is required to vaporize \( 2 \, \text{g} \) of water at \( 100^\circ \text{C} \) and \( 1 \, \text{atm} \)? (Latent heat = \( 2256 \, \text{J/g} \))

\( \Delta Q = m L \).

\( m = 2 \), \( L = 2256 \).

\( \Delta Q = 2 \times 2256 = 4512 \, \text{J} \).

4000 J
4512 J
5000 J
5500 J
2

What is the change in internal energy for \( 0.8 \, \text{moles} \) of an ideal gas heated from \( 310 \, \text{K} \) to \( 370 \, \text{K} \) at constant volume? (\( C_v = 20.8 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 0.8 \), \( C_v = 20.8 \), \( \Delta T = 370 - 310 = 60 \).

\( \Delta U = 0.8 \times 20.8 \times 60 = 998.4 \, \text{J} \approx 998 \, \text{J} \).

900 J
998 J
1050 J
1100 J
2

A diatomic gas undergoes an adiabatic expansion from \( 760 \, \text{K} \) to \( 380 \, \text{K} \) with \( 0.8 \, \text{moles} \). What is the work done? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \), \( \gamma = 1.4 \))

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

\( \mu = 0.8 \), \( R = 8.3 \), \( T_1 = 760 \), \( T_2 = 380 \), \( \gamma = 1.4 \).

\( W = \frac{0.8 \times 8.3 \times (760 - 380)}{1.4 - 1} = \frac{6.64 \times 380}{0.4} = 6312 \, \text{J} \).

6000 J
6312 J
6500 J
6800 J
2

A gas is compressed adiabatically, doing 300 J of work on the system. What is the change in internal energy?

For adiabatic (\( \Delta Q = 0 \)), First Law: \( \Delta U = -\Delta W \). Work on system: \( \Delta W = -300 \, \text{J} \) (negative by convention). \( \Delta U = -(-300) = 300 \, \text{J} \).

200 J
250 J
300 J
350 J
3

Which thermodynamic process involves a constant temperature?

An isothermal process is characterized by constant temperature (\( T = \text{constant} \)). For an ideal gas, this implies \( P V = \text{constant} \), with heat exchange balancing work done.

Adiabatic
Isobaric
Isochoric
Isothermal
4

A system absorbs 850 J of heat and has 300 J of work done on it. What is the change in internal energy?

First Law: \( \Delta Q = \Delta U + \Delta W \). \( \Delta Q = 850 \, \text{J} \), \( \Delta W = -300 \, \text{J} \) (work on system). \( 850 = \Delta U - 300 \Rightarrow \Delta U = 850 + 300 = 1150 \, \text{J} \).

1000 J
1150 J
1200 J
1250 J
2

In an isothermal process for an ideal gas, what happens to the internal energy?

For an ideal gas, internal energy (\( U \)) depends only on temperature. In an isothermal process, temperature remains constant (\( \Delta T = 0 \)), so \( \Delta U = 0 \).

Increases
Decreases
Remains constant
Becomes zero
3

0.5 kg of a substance at 15°C absorbs 1800 J of heat at constant volume, reaching 45°C. What is its specific heat capacity?

Specific heat: \( s = \frac{\Delta Q}{m \Delta T} \). \( \Delta Q = 1800 \, \text{J} \), \( m = 0.5 \, \text{kg} \), \( \Delta T = 45 - 15 = 30 \, \text{K} \). \( s = \frac{1800}{0.5 \times 30} = 120 \, \text{J kg}^{-1} \text{K}^{-1} \).

100 J kg⁻¹ K⁻¹
120 J kg⁻¹ K⁻¹
150 J kg⁻¹ K⁻¹
180 J kg⁻¹ K⁻¹
2

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