Thermodynamics Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A system absorbs 1200 J of heat and does 500 J of work on the surroundings at constant pressure. What is \(\Delta H\)?

At constant pressure, \(\Delta H = q_p\). Here, \(q = 1200 \, \text{J}\) (heat absorbed), so \(\Delta H = 1200 \, \text{J}\).

1200 J
700 J
1700 J
-700 J
1

Calculate \(\Delta H\) for \(CH_4(g) \rightarrow C(g) + 4H(g)\) using: \(\Delta H_f^\circ (CH_4,g) = -74.8 \, \text{kJ/mol}\), \(\Delta H_a (C,g) = 715 \, \text{kJ/mol}\), \(\Delta H_a (H,g) = 218 \, \text{kJ/mol}\).

\(\Delta H = [\Delta H_f^\circ (C) + 4 \times \Delta H_f^\circ (H)] - \Delta H_f^\circ (CH_4) = [715 + 4 \times 218] - (-74.8) = 715 + 872 + 74.8 = 1661.8 \, \text{kJ/mol}\).

789.8 kJ/mol
-1661.8 kJ/mol
1661.8 kJ/mol
74.8 kJ/mol
3

A system at constant pressure absorbs 1000 J of heat and expands by 0.4 L against 2.5 atm. What is \(\Delta U\)? (1 atm·L = 101.3 J)

\(w = -P\Delta V = -2.5 \times 0.4 \times 101.3 = -101.3 \, \text{J}\), \(q = 1000 \, \text{J}\), \(\Delta U = q + w = 1000 - 101.3 = 898.7 \, \text{J}\).

898.7 J
1101.3 J
1000 J
797.4 J
1

What is the change in internal energy (\( \Delta U \)) for the vaporization of 1 mol of water at 373 K if \( \Delta H = 40.79 \, \text{kJ/mol} \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \ce{H2O(l) \rightarrow H2O(g)} \), \( \Delta n_g = 1 \). Using \( \Delta H = \Delta U + \Delta n_g RT \), where \( RT = 8.314 \times 373 \times 10^{-3} = 3.1012 \, \text{kJ} \), we get \( \Delta U = \Delta H - \Delta n_g RT = 40.79 - 3.1012 = 37.69 \, \text{kJ/mol} \).

40.79 kJ/mol
37.69 kJ/mol
43.89 kJ/mol
34.60 kJ/mol
2

What is the work done when 2 mol of an ideal gas expands isothermally and reversibly from 5 L to 10 L at 300 K? (\(R = 8.314 \, \text{J/mol·K}\))

\(w = -nRT \ln(V_2/V_1) = -2 \times 8.314 \times 300 \times \ln(10/5) = -2 \times 2494.2 \times 0.693 = -3455.6 \, \text{J}\).

2494 J
4988 J
1728 J
-3456 J
4

Calculate \( \Delta S_{sys} \) when 1 mol of water vaporizes at 373 K. (\( \Delta H_{vap} = 40.79 \, \text{kJ/mol} \))

For \( \ce{H2O(l) \rightarrow H2O(g)} \) at 373 K, \( \Delta S_{sys} = \frac{\Delta H_{vap}}{T} = \frac{40.79 \times 10^3}{373} = 109.4 \, \text{J/K} \), representing the entropy increase due to phase change from liquid to gas.

40.79 J/K
80.00 J/K
109.4 J/K
150.0 J/K
3

In a process, 300 J of heat is absorbed by a system, and 100 J of work is done on the system. What is \(\Delta U\)?

Using the first law: \(\Delta U = q + w\). Here, \(q = +300 \, \text{J}\) (heat absorbed), \(w = +100 \, \text{J}\) (work done on system), so \(\Delta U = 300 + 100 = 400 \, \text{J}\).

400 J
200 J
-400 J
100 J
1

For an ideal gas (\(\gamma = 1.33\)) expanding adiabatically from 8 L to 16 L at 500 K, what is the final temperature?

\(T_2 = T_1 (V_1/V_2)^{\gamma-1} = 500 \times (8/16)^{0.33} = 500 \times (0.5)^{0.33} = 500 \times 0.793 = 396.5 \, \text{K}\).

250 K
396.5 K
630 K
500 K
2

Calculate \(\Delta G^\circ\) for a reaction with \(K = 2.5 \times 10^{-3}\) at 298 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta G^\circ = -RT \ln K = -8.314 \times 298 \times \ln(2.5 \times 10^{-3}) = -2477.6 \times (-5.991) = 14843 \, \text{J/mol} = 14.84 \, \text{kJ/mol}\).

-14.84 kJ/mol
5.99 kJ/mol
-7.42 kJ/mol
14.84 kJ/mol
4

A system absorbs 500 J of heat and performs 200 J of work. What is the change in internal energy (\(\Delta U\))?

Using the first law: \(\Delta U = q + w\). Here, \(q = +500 \, \text{J}\) (heat absorbed), \(w = -200 \, \text{J}\) (work done by system), so \(\Delta U = 500 - 200 = 300 \, \text{J}\).

300 J
700 J
-300 J
200 J
1

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