Thermodynamics Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following is always true for an isothermal process involving an ideal gas?

For an ideal gas in an isothermal process (\(\Delta T = 0\)), \(\Delta U = nC_v\Delta T = 0\), regardless of the path.

\(\Delta U = 0\)
\(\Delta H = 0\)
\(q = 0\)
\(w = 0\)
1

For the process \( \ce{CO(g) + 2H2(g) \rightarrow CH3OH(l)} \), \( \Delta H = -128 \, \text{kJ/mol} \) at 298 K, calculate \( \Delta U \). (\( R = 8.314 \, \text{J/mol·K} \))

For \( \ce{CO(g) + 2H2(g) \rightarrow CH3OH(l)} \), \( \Delta n_g = 0 - (1 + 2) = -3 \). Then, \( RT = 8.314 \times 298 \times 10^{-3} = 2.4776 \, \text{kJ} \). Using \( \Delta H = \Delta U + \Delta n_g RT \), we get \( \Delta U = \Delta H - \Delta n_g RT = -128 - (-3 \times 2.4776) = -128 + 7.4328 = -120.57 \, \text{kJ/mol} \).

-130.48 kJ/mol
-125.52 kJ/mol
-128.00 kJ/mol
-120.57 kJ/mol
4

Calculate \(\Delta S_{total}\) when 1 mol of ice melts at 263 K if \(\Delta H_{fus} = 6.01 \, \text{kJ/mol}\) and melting point is 273 K.

\(\Delta S_{sys} = \Delta H_{fus} / T_m = 6.01 \times 10^3 / 273 = 22.01 \, \text{J/K}\), \(\Delta S_{surr} = -\Delta H / T = -6.01 \times 10^3 / 263 = -22.85 \, \text{J/K}\), \(\Delta S_{total} = 22.01 - 22.85 = -0.84 \, \text{J/K}\).

22.01 J/K
-22.85 J/K
0 J/K
-0.84 J/K
4

During the isothermal reversible expansion of an ideal gas, the change in internal energy (\(\Delta U\)) is:

For an ideal gas, \(\Delta U = nC_v\Delta T\). In an isothermal process (\(\Delta T = 0\)), \(\Delta U = 0\).

Zero
Positive
Negative
Equal to work done
1

The bond enthalpy of \(O=O\) in \(O_2(g)\) is 498 kJ/mol. What is \(\Delta H_f\) of \(O(g)\)?

For \(O_2(g) \rightarrow 2O(g)\), \(\Delta H = 498 \, \text{kJ/mol}\). \(\Delta H_f (O_2) = 0\), so \(\Delta H = 2 \times \Delta H_f (O) - 0\), \(\Delta H_f (O) = 498 / 2 = 249 \, \text{kJ/mol}\).

498 kJ/mol
124.5 kJ/mol
249 kJ/mol
996 kJ/mol
3

What is the work done when 1 mol of an ideal gas expands isothermally and reversibly from 1 atm to 0.5 atm at 300 K? (\(R = 8.314 \, \text{J/mol·K}\))

\(w = -nRT \ln(P_1/P_2) = -1 \times 8.314 \times 300 \times \ln(1/0.5) = -2494.2 \times 0.693 = -1728 \, \text{J}\).

2494 J
0 J
-1728 J
-2494 J
3

Calculate the heat released when 60 g of ethanol is combusted if \(\Delta H_c = -1366.7 \, \text{kJ/mol}\). (Molar mass = 46 g/mol)

Moles = \(60 / 46 = 1.304 \, \text{mol}\), heat released = \(1.304 \times 1366.7 = 1782.2 \, \text{kJ}\).

1366.7 kJ
683.35 kJ
2733.4 kJ
1782.2 kJ
4

For a system at constant pressure, the heat absorbed is equal to:

At constant pressure, the heat absorbed \(q_p = \Delta H\), the change in enthalpy.

\(\Delta H\)
\(\Delta U\)
\(\Delta G\)
\(-\Delta U\)
1

For \(CO(g) + Cl_2(g) \rightarrow COCl_2(g)\), \(\Delta H = -108.3 \, \text{kJ/mol}\), \(\Delta S = -137 \, \text{J/K}\) at 298 K, what is \(\Delta G\)?

\(\Delta G = \Delta H - T\Delta S = -108.3 - 298 \times (-0.137) = -108.3 + 40.83 = -67.47 \, \text{kJ/mol}\).

108.3 kJ/mol
-40.83 kJ/mol
67.47 kJ/mol
-67.47 kJ/mol
4

A system releases 700 J of heat and performs 400 J of work on the surroundings. What is \(\Delta U\)?

\(\Delta U = q + w\). Here, \(q = -700 \, \text{J}\) (heat released), \(w = -400 \, \text{J}\) (work done by system), so \(\Delta U = -700 - 400 = -1100 \, \text{J}\).

-1100 J
1100 J
-300 J
300 J
1

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