Thermodynamics Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A system releases \( 790 \, \text{J} \) of heat and has \( 310 \, \text{J} \) of work done on it. What is the change in internal energy?

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

\( \Delta Q = -790 \) (heat released), \( \Delta W = -310 \) (work on system).

\( -790 = \Delta U - 310 \Rightarrow \Delta U = -790 + 310 = -480 \, \text{J} \).

-520 J
-480 J
-450 J
-400 J
2

What is the change in internal energy when \( 1 \, \text{mole} \) of an ideal gas is heated from \( 300 \, \text{K} \) to \( 350 \, \text{K} \) at constant volume? (\( C_v = 20.8 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 1 \), \( C_v = 20.8 \), \( \Delta T = 350 - 300 = 50 \).

\( \Delta U = 1 \times 20.8 \times 50 = 1040 \, \text{J} \).

800 J
1040 J
1200 J
1500 J
2

A system in a cyclic process absorbs \( 980 \, \text{J} \) of heat and performs \( 420 \, \text{J} \) of work. What is the heat rejected?

For cyclic: \( \Delta U = 0 \), \( Q_{\text{net}} = W \).

\( Q_{\text{absorb}} - Q_{\text{reject}} = W \).

\( 980 - Q_{\text{reject}} = 420 \Rightarrow Q_{\text{reject}} = 980 - 420 = 560 \, \text{J} \).

520 J
560 J
600 J
640 J
2

A system absorbs \( 920 \, \text{J} \) of heat and performs \( 280 \, \text{J} \) of work. What is the change in internal energy?

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

\( \Delta Q = 920 \), \( \Delta W = 280 \) (work by system).

\( 920 = \Delta U + 280 \Rightarrow \Delta U = 920 - 280 = 640 \, \text{J} \).

600 J
640 J
700 J
750 J
2

What is the primary implication of the First Law of Thermodynamics?

The First Law of Thermodynamics is a statement of energy conservation: \( \Delta Q = \Delta U + \Delta W \). It implies that the total energy supplied to a system (as heat) equals the increase in internal energy plus the work done by the system.

Heat engines can achieve 100% efficiency
Energy cannot be created or destroyed
Heat flows from cold to hot spontaneously
Entropy always decreases
2

How many calories are equivalent to \( 2093 \, \text{J} \) of heat? (1 cal = \( 4.186 \, \text{J} \))

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

\( \frac{2093}{4.186} \approx 500 \, \text{cal} \).

450 cal
500 cal
550 cal
600 cal
2

A gas at 4 atm and 300 K in a 5 L container is compressed isothermally to 2 L. What is the work done on the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

Isothermal: \( W = \mu R T \ln\left(\frac{V_2}{V_1}\right) \). \( P_1 V_1 = \mu R T \Rightarrow 4 \times 5 = \mu \times 8.3 \times 300 \Rightarrow \mu = \frac{20}{2490} \approx 0.008 \, \text{mol} \). \( W = 0.008 \times 8.3 \times 300 \times \ln\left(\frac{2}{5}\right) = 19.92 \times (-0.916) \approx -18.25 \, \text{J} \) (work by gas negative). Work on gas = \( 18.25 \, \text{J} \approx 18 \, \text{J} \).

15 J
18 J
20 J
25 J
2

A diatomic gas undergoes an adiabatic expansion from \( 860 \, \text{K} \) to \( 430 \, \text{K} \) with \( 0.5 \, \text{moles} \). What is the work done? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \), \( \gamma = 1.4 \))

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

\( \mu = 0.5 \), \( R = 8.3 \), \( T_1 = 860 \), \( T_2 = 430 \), \( \gamma = 1.4 \).

\( W = \frac{0.5 \times 8.3 \times (860 - 430)}{1.4 - 1} = \frac{4.15 \times 430}{0.4} = 4467.5 \, \text{J} \approx 4468 \, \text{J} \).

4200 J
4468 J
4600 J
4800 J
2

An ideal gas expands isothermally at \( 540 \, \text{K} \) from \( 10 \, \text{L} \) to \( 30 \, \text{L} \) with \( 0.3 \, \text{moles} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 0.3 \), \( R = 8.3 \), \( T = 540 \), \( V_2 = 30 \), \( V_1 = 10 \).

\( W = 0.3 \times 8.3 \times 540 \times \ln\left(\frac{30}{10}\right) = 1344.6 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 1344.6 \times 1.0986 \approx 1477 \, \text{J} \).

1300 J
1477 J
1600 J
1700 J
2

A system absorbs \( 800 \, \text{J} \) of heat and has \( 350 \, \text{J} \) of work done on it. What is the change in internal energy?

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

\( \Delta Q = 800 \), \( \Delta W = -350 \) (work done on system).

\( 800 = \Delta U - 350 \Rightarrow \Delta U = 800 + 350 = 1150 \, \text{J} \).

1000 J
1150 J
1200 J
1300 J
2

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