Thermodynamics Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Calculate the heat absorbed when 32 g of sulfur is heated from 30°C to 130°C at constant pressure. (\(C_p = 22.7 \, \text{J/mol·K}\), molar mass = 32 g/mol)

Moles = \(32 / 32 = 1 \, \text{mol}\), \(\Delta T = 130 - 30 = 100 \, \text{K}\), \(q_p = nC_p\Delta T = 1 \times 22.7 \times 100 = 2270 \, \text{J} = 2.27 \, \text{kJ}\).

1.135 kJ
4.54 kJ
22.7 kJ
2.27 kJ
4

Calculate \( \Delta S_{sys} \) for the condensation of 1 mol of ethanol at 351 K. (\( \Delta H_{vap} = 38.6 \, \text{kJ/mol} \))

For \( C_2H_5OH(g) \rightarrow C_2H_5OH(l) \), \( \Delta S_{sys} = -\frac{\Delta H_{vap}}{T} = -\frac{38.6 \times 10^3}{351} \approx -110.0 \, \text{J/K·mol} \), as condensation decreases entropy.

38.6 J/K·mol
110.0 J/K·mol
-110.0 J/K·mol
-351.0 J/K·mol
3

For which process does entropy decrease?

Entropy decreases when disorder decreases, e.g., \(H_2O(l) \rightarrow H_2O(s)\), as liquid to solid reduces randomness.

\(H_2(g) \rightarrow 2H(g)\)
\(NaHCO_3(s) \rightarrow Na_2CO_3(s) + CO_2(g) + H_2O(g)\)
\(H_2O(l) \rightarrow H_2O(s)\)
Temperature rise from 0 K to 100 K
3

For an ideal gas undergoing adiabatic expansion from 2 L to 4 L at 400 K (\( \gamma = 1.67 \)), what is the final temperature?

For adiabatic process, \( T_2 = T_1 (V_1/V_2)^{\gamma-1} = 400 \times (2/4)^{0.67} \approx 400 \times 0.62996 = 252.0 \, \text{K} \).

200 K
252 K
315 K
400 K
2

Calculate \(\Delta S_{surr}\) when 1 mol of \(H_2O(l)\) forms under standard conditions at 298 K, given \(\Delta_f H^\circ = -286 \, \text{kJ/mol}\).

\(\Delta S_{surr} = -\Delta H / T = -(-286 \times 10^3) / 298 = 959.7 \, \text{J/K·mol}\).

-959.7 J/K·mol
286 J/K·mol
-286 J/K·mol
959.7 J/K·mol
4

Calculate the work done when 1 mol of an ideal gas (\( \gamma = 1.67 \)) is compressed adiabatically from 20 L to 4 L at 300 K. (\( R = 8.314 \, \text{J/mol·K} \))

For adiabatic compression, \( T_2 = T_1 (V_1/V_2)^{\gamma-1} = 300 \times (20/4)^{0.67} \approx 300 \times 2.63 = 789.0 \, \text{K} \). Then, \( C_v = R/(\gamma-1) = 8.314 / 0.67 \approx 12.41 \, \text{J/mol·K} \). Work done: \( w = nC_v\Delta T = 1 \times 12.41 \times (789.0 - 300) \approx 6068.3 \, \text{J} \).

0 J
-6068.3 J
6068.3 J
3034.2 J
3

For a reversible isothermal expansion of an ideal gas, the work done is given by:

For reversible isothermal expansion, \(w = -nRT \ln(V_2/V_1)\), where \(V_2 > V_1\).

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

Calculate \(\Delta H\) for \(C_2H_5OH(l) \rightarrow C_2H_4(g) + H_2O(l)\) using: \(\Delta H_f^\circ (C_2H_5OH,l) = -277.7 \, \text{kJ/mol}\), \(\Delta H_f^\circ (C_2H_4,g) = 52.3 \, \text{kJ/mol}\), \(\Delta H_f^\circ (H_2O,l) = -285.8 \, \text{kJ/mol}\).

\(\Delta H = [\Delta H_f^\circ (C_2H_4) + \Delta H_f^\circ (H_2O)] - \Delta H_f^\circ (C_2H_5OH) = [52.3 + (-285.8)] - (-277.7) = -233.5 + 277.7 = 44.2 \, \text{kJ/mol}\).

-44.2 kJ/mol
277.7 kJ/mol
44.2 kJ/mol
615.8 kJ/mol
3

Calculate \(\Delta S_{surr}\) when 1 mol of a substance melts at 300 K with \(\Delta H_{fus} = 9 \, \text{kJ/mol}\).

\(\Delta S_{surr} = -\Delta H / T = -9 \times 10^3 / 300 = -30 \, \text{J/K·mol}\).

30 J/K·mol
9 J/K·mol
300 J/K·mol
-30 J/K·mol
4

Using bond enthalpies (\(C-H = 413 \, \text{kJ/mol}\), \(Cl-Cl = 243 \, \text{kJ/mol}\), \(C-Cl = 328 \, \text{kJ/mol}\), \(H-Cl = 431 \, \text{kJ/mol}\)), calculate \(\Delta H\) for \(CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)\).

Bonds broken: \(C-H (413) + Cl-Cl (243) = 656 \, \text{kJ}\). Bonds formed: \(C-Cl (328) + H-Cl (431) = 759 \, \text{kJ}\). \(\Delta H = 656 - 759 = -103 \, \text{kJ/mol}\).

103 kJ/mol
-656 kJ/mol
-103 kJ/mol
759 kJ/mol

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