Thermodynamics Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For a chemical reaction carried out at constant temperature (ideal-gas approximation) with \( \Delta U = 250 \,\text{J} \) and no change in moles of gas (\( \Delta n_g = 0 \)), what is the change in enthalpy \( (\Delta H) \)?

For reactions at constant temperature (ideal gas), \( \Delta H = \Delta U + \Delta n_g RT \). With \( \Delta n_g = 0 \), \( \Delta H = \Delta U = 250 \,\text{J} \).

250 J
0 J
500 J
-250 J
1

For the reaction \(N_2(g) + O_2(g) \rightarrow 2NO(g)\), \(\Delta H = 180.6 \, \text{kJ/mol}\) at 298 K. What is \(\Delta U\) if the gas behaves ideally? (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta n_g = 2 - 2 = 0\), \(\Delta H = \Delta U + \Delta n_g RT\). Since \(\Delta n_g = 0\), \(\Delta U = \Delta H = 180.6 \, \text{kJ/mol}\).

180.6 kJ/mol
183.1 kJ/mol
178.1 kJ/mol
90.3 kJ/mol
1

For the reaction \(H_2(g) + I_2(g) \rightarrow 2HI(g)\), \(\Delta H = 52.96 \, \text{kJ/mol}\) at 298 K. What is \(\Delta U\)? (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta n_g = 2 - 2 = 0\), so \(\Delta H = \Delta U + \Delta n_g RT = \Delta U\). Thus, \(\Delta U = 52.96 \, \text{kJ/mol}\).

50.48 kJ/mol
55.44 kJ/mol
52.96 kJ/mol
0 kJ/mol
3

What is the entropy change when 1 mol of an ideal gas is compressed isothermally from 10 L to 5 L? (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta S = nR \ln(V_2/V_1) = 1 \times 8.314 \times \ln(5/10) = 8.314 \times (-0.693) = -5.76 \, \text{J/K}\).

5.76 J/K
8.314 J/K
2.88 J/K
-5.76 J/K
4

For the combustion of 16 g of methane at 298 K (\( \Delta H_c = -890.30 \, \text{kJ/mol} \), molar mass = 16 g/mol), what is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

For \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \), \( \Delta n_g = 1 - 3 = -2 \). Then, \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -890.30 - (-2 \times 2.4776) = -890.30 + 4.9552 = -885.34 \, \text{kJ/mol} \).

-885.3 kJ/mol
-890.3 kJ/mol
-895.3 kJ/mol
-880.3 kJ/mol
1

For the reaction \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\) at 298 K. What is \(\Delta U\)? (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta n_g = 1 - 1 = 0\), \(RT = 2.48 \, \text{kJ}\), \(\Delta H = \Delta U + \Delta n_g RT\), \(\Delta U = -393.5 - 0 = -393.5 \, \text{kJ/mol}\).

-393.5 kJ/mol
-391 kJ/mol
-396 kJ/mol
-386 kJ/mol
1

In an isobaric process, 600 J of heat is supplied, and the system does 200 J of work. What is \(\Delta U\)?

At constant pressure, \(\Delta H = q_p = 600 \, \text{J}\), \(w = -200 \, \text{J}\) (work done by system). \(\Delta U = q + w = 600 - 200 = 400 \, \text{J}\).

400 J
800 J
200 J
-400 J
1

Calculate \(\Delta G^\circ\) for a reaction with \(K = 10^4\) at 298 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta G^\circ = -RT \ln K = -8.314 \times 298 \times \ln(10^4) = -8.314 \times 298 \times 9.21 = -22823 \, \text{J/mol} = -22.82 \, \text{kJ/mol}\).

22.82 kJ/mol
11.41 kJ/mol
45.64 kJ/mol
-22.82 kJ/mol
4

For a reversible isothermal compression of an ideal gas, the heat absorbed is:

For an isothermal process (\(\Delta U = 0\)), \(\Delta U = q + w = 0\), so \(q = -w\). For reversible compression, \(w = -nRT \ln(V_2/V_1)\) (positive), so \(q = nRT \ln(V_2/V_1)\), which is negative since \(V_2 < V_1\).

\(nRT \ln(V_1/V_2)\)
\(-nRT \ln(V_2/V_1)\)
Zero
\(nRT\)
2

For an ideal gas (\(\gamma = 1.33\)) compressed adiabatically from 18 L to 6 L at 300 K, what is the final temperature?

\(T_2 = T_1 (V_1/V_2)^{\gamma-1} = 300 \times (18/6)^{0.33} = 300 \times 3^{0.33} = 300 \times 1.44 = 432 \, \text{K}\).

100 K
288 K
432 K
900 K
3

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