Question

A mixture of gas expands from \(0.03 \text{ m}^3\) to \(0.06 \text{ m}^3\) at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is

• A. \(30 \text{ kJ} \)
• B. \(54 \text{ kJ} \)
• C. \(114 \text{ kJ} \)
• D. \(84 \text{ kJ} \)

Hint:

Keep unit consistency intact! Work done at constant pressure can be quickly found by multiplying pressure in MPa directly with volume in $\text{m}^3$, which gives work in MJ: $W = 1 \text{ MPa} \times 0.03 \text{ m}^3 = 0.03 \text{ MJ} = 30 \text{ kJ}$. Then, $\Delta U = 84 - 30 = 54 \text{ kJ}$.

Answer 1

The Correct Option is B

Concept: According to the First Law of Thermodynamics for a closed system undergoing a non-cyclic process, the net heat interactions ($Q$) equal the sum of the change in internal energy ($\Delta U$) and the work performed ($W$): \[ Q = \Delta U + W \] For a quasi-static displacement process operating under constant pressure conditions (isobaric process), the boundary work output ($W$) is given by the integral of pressure over volume: \[ W = \int P \, dV = P(V_2 - V_1) \] By evaluating the boundary work and employing the first law, we can isolate the net internal energy alteration.

Step 1: Extract parameters and equalize metric prefixes to standard SI base units.
The parameters given in the problem statement are:

• Constant Pressure, \(P = 1 \text{ MPa} = 1 \times 10^6 \text{ N/m}^2\)

• Initial Volume, \(V_1 = 0.03 \text{ m}^3\)

• Final Volume, \(V_2 = 0.06 \text{ m}^3\)

• Heat absorbed by the system, \(Q = +84 \text{ kJ} = 84 \times 10^3 \text{ J}\)

Step 2: Calculate the boundary work performed during expansion.
Using the constant-pressure work equation: \[ W = P \times (V_2 - V_1) \] Substitute the values: \[ W = (1 \times 10^6 \text{ Pa}) \times (0.06 \text{ m}^3 - 0.03 \text{ m}^3) \] \[ W = 1 \times 10^6 \times 0.03 = 30000 \text{ J} \] Converting Joules to kilojoules gives: \[ W = \frac{30000}{1000} \text{ kJ} = 30 \text{ kJ} \]

Step 3: Solve for the change in internal energy (\(\Delta U\)) via the First Law.
Rearranging the first law equation to solve for $\Delta U$: \[ \Delta U = Q - W \] Substituting our calculated heat interaction and boundary work values: \[ \Delta U = 84 \text{ kJ} - 30 \text{ kJ} = 54 \text{ kJ} \] Thus, the change in internal energy of the gas mixture is equal to 54 kJ, corresponding to Option (B).