Thermodynamics Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the change in internal energy for \( 0.4 \, \text{moles} \) of an ideal gas heated from \( 290 \, \text{K} \) to \( 330 \, \text{K} \) at constant volume? (\( C_v = 20.8 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 0.4 \), \( C_v = 20.8 \), \( \Delta T = 330 - 290 = 40 \).

\( \Delta U = 0.4 \times 20.8 \times 40 = 332.8 \, \text{J} \approx 333 \, \text{J} \).

300 J
333 J
350 J
400 J
2

An ideal gas expands isothermally at \( 570 \, \text{K} \) from \( 12 \, \text{L} \) to \( 36 \, \text{L} \) with \( 0.4 \, \text{moles} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 0.4 \), \( R = 8.3 \), \( T = 570 \), \( V_2 = 36 \), \( V_1 = 12 \).

\( W = 0.4 \times 8.3 \times 570 \times \ln\left(\frac{36}{12}\right) = 1892.4 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 1892.4 \times 1.0986 \approx 2079 \, \text{J} \).

1900 J
2079 J
2200 J
2300 J
2

What is the change in internal energy for \( 0.5 \, \text{moles} \) of an ideal gas heated from \( 260 \, \text{K} \) to \( 310 \, \text{K} \) at constant volume? (\( C_v = 20.8 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 0.5 \), \( C_v = 20.8 \), \( \Delta T = 310 - 260 = 50 \).

\( \Delta U = 0.5 \times 20.8 \times 50 = 520 \, \text{J} \).

480 J
520 J
550 J
600 J
2

Which condition is essential for a process to be considered reversible?

A reversible process requires the absence of dissipative effects (e.g., friction, viscosity), allowing the system and surroundings to return to their original states. Quasi-static conditions alone are insufficient without eliminating dissipation.

Rapid temperature changes
Presence of friction
No dissipative effects
Constant volume
3

Why does the First Law of Thermodynamics allow some processes that the Second Law prohibits?

The First Law ensures energy conservation (\( \Delta Q = \Delta U + \Delta W \)), permitting any energy-balanced process. The Second Law introduces directionality and efficiency limits (e.g., no 100% heat-to-work conversion), restricting feasible processes.

It ignores temperature
It only conserves energy
It ensures reversibility
It defines equilibrium
2

Which of the following correctly describes the concept of heat?

Heat is energy transferred due to a temperature difference, not a stored property or work form. Option B is correct.

It is a state variable
It is energy transferred due to temperature difference
It is mechanical work done
It remains constant in all processes
2

A system absorbs \( 450 \, \text{J} \) of heat and performs \( 150 \, \text{J} \) of work. What is the change in internal energy?

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

\( \Delta Q = 450 \), \( \Delta W = 150 \).

\( 450 = \Delta U + 150 \Rightarrow \Delta U = 450 - 150 = 300 \, \text{J} \).

250 J
300 J
350 J
400 J
2

A system absorbs \( 500 \, \text{J} \) of heat and does \( 200 \, \text{J} \) of work. What is the change in its internal energy?

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

Given: \( \Delta Q = 500 \, \text{J} \), \( \Delta W = 200 \, \text{J} \).

\( 500 = \Delta U + 200 \).

\( \Delta U = 500 - 200 = 300 \, \text{J} \).

200 J
300 J
400 J
500 J
2

Which property of an ideal gas simplifies its internal energy calculation?

For an ideal gas, internal energy (\( U \)) depends solely on temperature because intermolecular forces are negligible, reducing \( U \) to the sum of molecular kinetic energies, independent of pressure or volume interactions.

Constant pressure
Negligible intermolecular forces
Fixed volume
High density
2

What does the Clausius statement of the Second Law imply?

The Clausius statement states that heat cannot flow spontaneously from a colder object to a hotter object without external work. This establishes a direction for natural heat transfer processes.

Heat engines are 100% efficient
Heat flows from cold to hot without work
Heat cannot flow from cold to hot without work
Work is always converted to heat
3

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