Thermodynamics Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A system releases \( 670 \, \text{J} \) of heat and has \( 230 \, \text{J} \) of work done on it. What is the change in internal energy?

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

\( \Delta Q = -670 \) (heat released), \( \Delta W = -230 \) (work on system).

\( -670 = \Delta U - 230 \Rightarrow \Delta U = -670 + 230 = -440 \, \text{J} \).

-500 J
-440 J
-400 J
-350 J
2

0.2 moles of an ideal gas at 360 K are compressed isothermally from 8 L to 2 L. What is the heat released? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

Isothermal: \( \Delta U = 0 \), \( \Delta Q = \Delta W \). \( W = \mu R T \ln\left(\frac{V_2}{V_1}\right) \). \( \mu = 0.2 \), \( T = 360 \), \( V_2 = 2 \), \( V_1 = 8 \). \( W = 0.2 \times 8.3 \times 360 \times \ln\left(\frac{2}{8}\right) = 597.6 \times (-1.386) \approx -829 \, \text{J} \) (work by gas). Heat released = \( 829 \, \text{J} \).

800 J
829 J
850 J
900 J
2

Which statement best describes the Second Law of Thermodynamics?

The Second Law states that not all processes allowed by the First Law (energy conservation) occur naturally. It limits efficiency (Kelvin-Planck) and direction of heat flow (Clausius), disallowing 100% conversion of heat to work or spontaneous heat flow from cold to hot.

Energy is conserved in all processes
All heat can be converted into work
Some processes are inherently irreversible
Temperature is absolute
3

What thermodynamic principle explains why a bullet’s high speed does not increase its temperature?

Internal energy in thermodynamics relates to the random molecular motion, not the macroscopic kinetic energy of the system as a whole. A bullet’s high speed is bulk motion, not affecting molecular energy or temperature.

Heat equals work
Internal energy excludes bulk motion
Temperature depends on pressure
Energy is lost to friction
2

A gas is compressed isothermally at \( 320 \, \text{K} \) releasing \( 960 \, \text{J} \) of heat. What is the work done on the gas? (\( \Delta U = 0 \))

For isothermal: \( \Delta U = 0 \), \( \Delta Q = \Delta W \).

\( \Delta Q = -960 \, \text{J} \) (heat released).

\( \Delta W = -960 \, \text{J} \) (work by gas negative), so work on gas = \( 960 \, \text{J} \).

800 J
960 J
1000 J
1200 J
2

A gas expands adiabatically from \( 9 \, \text{atm} \) and \( 6 \, \text{L} \) to \( 3 \, \text{atm} \). What is the final volume? (\( \gamma = 1.4 \))

\( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 9 \times 6^{1.4} = 3 \times V_2^{1.4} \).

\( V_2^{1.4} = \frac{9}{3} \times 6^{1.4} = 3 \times 6^{1.4} \).

\( 6^{1.4} \approx 12.29 \), \( V_2^{1.4} = 3 \times 12.29 \approx 36.87 \).

\( V_2 = (36.87)^{1/1.4} \approx 10.9 \, \text{L} \).

9 L
10.9 L
12 L
14 L
2

A system absorbs \( 850 \, \text{J} \) of heat and has \( 400 \, \text{J} \) of work done on it. What is the change in internal energy?

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

\( \Delta Q = 850 \), \( \Delta W = -400 \) (work done on system).

\( 850 = \Delta U - 400 \Rightarrow \Delta U = 850 + 400 = 1250 \, \text{J} \).

1000 J
1250 J
1500 J
1750 J
2

An ideal gas expands isothermally at \( 510 \, \text{K} \) from \( 8 \, \text{L} \) to \( 24 \, \text{L} \) with \( 0.2 \, \text{moles} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

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

\( \mu = 0.2 \), \( R = 8.3 \), \( T = 510 \), \( V_2 = 24 \), \( V_1 = 8 \).

\( W = 0.2 \times 8.3 \times 510 \times \ln\left(\frac{24}{8}\right) = 846.6 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 846.6 \times 1.0986 \approx 930 \, \text{J} \).

850 J
930 J
1000 J
1100 J
2

A system absorbs 600 J of heat and does 150 J of work. What is the change in internal energy?

First Law: \( \Delta Q = \Delta U + \Delta W \). Given \( \Delta Q = 600 \, \text{J} \), \( \Delta W = 150 \, \text{J} \) (work by system). \( 600 = \Delta U + 150 \Rightarrow \Delta U = 600 - 150 = 450 \, \text{J} \).

400 J
450 J
500 J
550 J
2

A monatomic gas undergoes an adiabatic expansion from \( 820 \, \text{K} \) to \( 410 \, \text{K} \) with \( 0.9 \, \text{moles} \). What is the work done? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \), \( \gamma = 1.67 \))

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

\( \mu = 0.9 \), \( R = 8.3 \), \( T_1 = 820 \), \( T_2 = 410 \), \( \gamma = 1.67 \).

\( W = \frac{0.9 \times 8.3 \times (820 - 410)}{1.67 - 1} = \frac{7.47 \times 410}{0.67} \approx 4570.15 \, \text{J} \approx 4570 \, \text{J} \).

4300 J
4570 J
4700 J
4900 J
2

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