Thermodynamics Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A system at constant volume absorbs 208 J of heat, and its temperature increases from 300 K to 310 K. What is \( C_v \)? (1 mol diatomic gas)

For a diatomic ideal gas at constant volume, \( \Delta U = q_v = nC_v\Delta T \). Given \( q_v = 208 \, \text{J} \), \( n = 1 \, \text{mol} \), \( \Delta T = 310 - 300 = 10 \, \text{K} \), we get \( 208 = 1 \times C_v \times 10 \), so \( C_v = 208 / 10 = 20.8 \, \text{J/mol·K} \), typical for a diatomic gas (\( C_v \approx 2.5R \)).

12.5 J/mol·K
20.8 J/mol·K
30.0 J/mol·K
90.0 J/mol·K
2

Calculate \(\Delta H\) for \(C(s) + 2H_2O(g) \rightarrow CO_2(g) + 2H_2(g)\) using: \(\Delta H_f^\circ (H_2O,g) = -241.8 \, \text{kJ/mol}\), \(\Delta H_f^\circ (CO_2,g) = -393.5 \, \text{kJ/mol}\).

\(\Delta H = [\Delta H_f^\circ (CO_2) + 2 \times \Delta H_f^\circ (H_2)] - [\Delta H_f^\circ (C) + 2 \times \Delta H_f^\circ (H_2O)] = [-393.5 + 0] - [0 + 2(-241.8)] = -393.5 + 483.6 = 90.1 \, \text{kJ/mol}\).

-90.1 kJ/mol
483.6 kJ/mol
90.1 kJ/mol
-393.5 kJ/mol
3

Calculate the work done when 1 mol of an ideal gas expands isothermally and reversibly from 5 atm to 1 atm at 300 K. (\(R = 8.314 \, \text{J/mol·K}\))

\(w = -nRT \ln(P_1/P_2) = -1 \times 8.314 \times 300 \times \ln(5/1) = -2494.2 \times 1.609 = -4013 \, \text{J}\).

2494 J
4013 J
-4013 J
-2494 J
3

The enthalpy of combustion of glucose is -2808 kJ/mol. How much heat is released when 90 g of glucose is burnt? (Molar mass of \(C_6H_{12}O_6 = 180 \, \text{g/mol}\))

Moles = \(90 / 180 = 0.5 \, \text{mol}\), heat released = \(0.5 \times 2808 = 1404 \, \text{kJ}\).

2808 kJ
702 kJ
5616 kJ
1404 kJ
4

The standard enthalpy change for the reaction \(N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\) is given by:

The standard enthalpy change for a reaction forming a compound from its elements in their standard states is the enthalpy of formation (\(\Delta_f H^\circ\)) multiplied by the moles formed. For \(2NH_3\), \(\Delta H = 2 \times \Delta_f H^\circ (NH_3)\).

\(2 \times \Delta_f H^\circ (NH_3)\)
\(\Delta_f H^\circ (N_2) + 3 \times \Delta_f H^\circ (H_2)\)
\(\Delta_f H^\circ (NH_3) - \Delta_f H^\circ (N_2)\)
\(\Delta_f H^\circ (NH_3) / 2\)
1

Calculate \(\Delta S_{surr}\) when 2 mol of water freezes at 273 K if \(\Delta H_{fus} = 6.01 \, \text{kJ/mol}\).

For freezing, \(\Delta H = -n \times \Delta H_{fus} = -2 \times 6.01 = -12.02 \, \text{kJ}\), \(\Delta S_{surr} = -\Delta H / T = -(-12.02 \times 10^3) / 273 = 44.03 \, \text{J/K}\).

-44.03 J/K
22.01 J/K
6.01 J/K
44.03 J/K
4

Calculate the heat required to raise the temperature of 40 g of water from 25°C to 35°C. (Specific heat of water = 4.18 J/g·K)

\(q = m \times C \times \Delta T = 40 \times 4.18 \times 10 = 1672 \, \text{J} = 1.672 \, \text{kJ}\).

0.836 kJ
4.18 kJ
3.344 kJ
1.672 kJ
4

The equilibrium constant for a reaction is 100 at 300 K. Calculate \(\Delta G^\circ\). (\(R = 8.314 \, \text{J/mol·K}\))

\(\Delta G^\circ = -RT \ln K = -8.314 \times 300 \times \ln(100) = -8.314 \times 300 \times 4.605 = -11488 \, \text{J/mol} = -11.49 \, \text{kJ/mol}\).

-5.74 kJ/mol
11.49 kJ/mol
-22.98 kJ/mol
-11.49 kJ/mol
4

Calculate \(\Delta H\) for \(C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)\) given: \(C(s) + O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -393.5 \, \text{kJ/mol}\); \(CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)\), \(\Delta H = -283 \, \text{kJ/mol}\).

Using Hess’s law: \(\Delta H = -393.5 - (-283) = -393.5 + 283 = -110.5 \, \text{kJ/mol}\).

-676.5 kJ/mol
283 kJ/mol
-110.5 kJ/mol
393.5 kJ/mol
3

For an ideal gas undergoing reversible adiabatic compression, which relationship holds?

In a reversible adiabatic process, \(P^{1-\gamma} T^\gamma = \text{constant}\), derived from \(PV^\gamma = \text{constant}\) and the ideal gas law.

\(P^{1-\gamma} T^\gamma = \text{constant}\)
\(PV = \text{constant}\)
\(T/V = \text{constant}\)
\(P/T = \text{constant}\)
1

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