Thermodynamics Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Calculate the heat required to raise the temperature of 27 g of Al from 25°C to 45°C. (Molar heat capacity of Al = 24 J/mol·K, molar mass = 27 g/mol)

Moles = \(27 / 27 = 1 \, \text{mol}\), \(\Delta T = 20 \, \text{K}\), \(q = nC_p\Delta T = 1 \times 24 \times 20 = 480 \, \text{J} = 0.48 \, \text{kJ}\).

0.24 kJ
0.96 kJ
0.72 kJ
0.48 kJ
4

The standard enthalpy of formation of \(SO_3(g)\) is -395.7 kJ/mol. What is \(\Delta H\) for \(2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)\) if \(\Delta H_f^\circ (SO_2,g) = -296.8 \, \text{kJ/mol}\)?

\(\Delta H = [2 \times \Delta H_f^\circ (SO_3)] - [2 \times \Delta H_f^\circ (SO_2) + \Delta H_f^\circ (O_2)] = [2 \times (-395.7)] - [2 \times (-296.8) + 0] = -791.4 + 593.6 = -197.8 \, \text{kJ/mol}\).

-197.8 kJ/mol
-98.9 kJ/mol
197.8 kJ/mol
-395.7 kJ/mol
1

Calculate the entropy change (\(\Delta S\)) when 1 mol of an ideal gas is compressed isothermally from 20 L to 5 L at 400 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta S = nR \ln(V_2/V_1) = 1 \times 8.314 \times \ln(5/20) = 8.314 \times \ln(0.25) = 8.314 \times (-1.386) = -11.52 \, \text{J/K}\).

11.52 J/K
-11.52 J/K
5.76 J/K
-23.04 J/K
2

For the reaction \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \), \( \Delta H = -286.00 \, \text{kJ/mol} \). If \( \Delta n_g = -1.5 \), what is \( \Delta U \) at 298 K? (\( R = 8.314 \, \text{J/mol·K} \))

For \( \Delta n_g = -1.5 \), \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -286.00 - (-1.5 \times 2.4776) = -286.00 + 3.7164 = -282.28 \, \text{kJ/mol} \).

-286 kJ/mol
-282.3 kJ/mol
-289.7 kJ/mol
-280 kJ/mol
2

Calculate \(\Delta H\) for \(2C(s) + O_2(g) \rightarrow 2CO(g)\) given: \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(2CO(g) + O_2(g) \rightarrow 2CO_2(g)\), \(\Delta H = -566 \, \text{kJ/mol}\).

Multiply first equation by 2: \(2C(s) + 2O_2(g) \rightarrow 2CO_2(g)\), \(\Delta H = -787 \, \text{kJ}\). Reverse second: \(2CO_2(g) \rightarrow 2CO(g) + O_2(g)\), \(\Delta H = 566 \, \text{kJ}\). Add: \(\Delta H = -787 + 566 = -221 \, \text{kJ}\).

-959.5 kJ
566 kJ
-221 kJ
393.5 kJ
3

For an ideal gas undergoing reversible adiabatic expansion, which quantity remains constant?

In a reversible adiabatic process, \(PV^\gamma = \text{constant}\), where \(\gamma = C_p/C_v\).

\(P\)
\(PV^\gamma\)
\(T\)
\(V\)
2

Calculate \(\Delta H\) for \(2C(s) + H_2(g) \rightarrow C_2H_2(g)\) given: \(C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l)\), \(\Delta H = -1299.5 \, \text{kJ/mol}\); \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\), \(\Delta H = -285.8 \, \text{kJ/mol}\).

Reverse first: \(\Delta H = 1299.5 \, \text{kJ}\). Second (×2): \(\Delta H = -787 \, \text{kJ}\). Third: \(\Delta H = -285.8 \, \text{kJ}\). Add: \(1299.5 - 787 - 285.8 = 226.7 \, \text{kJ/mol}\).

-226.7 kJ/mol
226.7 kJ/mol
1299.5 kJ/mol
-1072.8 kJ/mol
2

For the reaction \( 2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g) \), \( \Delta n_g = -3 \). If \( \Delta H = -1124.00 \, \text{kJ/mol} \) at 298 K, what is \( \Delta U \)? (\( R = 8.314 \, \text{J/mol·K} \))

\( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ/mol} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), \( \Delta U = -1124.00 - (-3 \times 2.4776) = -1124.00 + 7.4328 = -1116.57 \, \text{kJ/mol} \).

-1116.6 kJ/mol
-1121.5 kJ/mol
-1119.0 kJ/mol
-1124.0 kJ/mol
1

Calculate \(\Delta S_{total}\) when 1 mol of a liquid vaporizes at 400 K if \(\Delta H_{vap} = 35 \, \text{kJ/mol}\).

\(\Delta S_{sys} = \Delta H_{vap} / T = 35 \times 10^3 / 400 = 87.5 \, \text{J/K}\), \(\Delta S_{surr} = -\Delta H / T = -35 \times 10^3 / 400 = -87.5 \, \text{J/K}\), \(\Delta S_{total} = 87.5 - 87.5 = 0 \, \text{J/K}\).

87.5 J/K
-87.5 J/K
35 J/K
0 J/K
4

A system absorbs 800 J of heat at constant volume, and its temperature rises by 10 K. If \(C_v = 20 \, \text{J/mol·K}\), how many moles are present?

At constant volume, \(\Delta U = q_v = nC_v\Delta T\). So, \(800 = n \times 20 \times 10\), \(n = 800 / 200 = 4 \, \text{mol}\).

4 mol
2 mol
8 mol
40 mol
1

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