Question:

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from \( P_1 \) to \( P_2 \) is \( \alpha \) Joule \((P_1 = 21.7 \text{ Pa}, P_2 = 30 \text{ Pa}, C_v = 21 \text{ J/K mol}, R = 8.3 \text{ J/mol K})\). The value of \( \alpha \) is ________.

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For constant volume processes, heat supplied equals change in internal energy.
Updated On: Feb 5, 2026
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The Correct Option is B

Solution and Explanation

Step 1: Identify the nature of the process.
From the \(P\)-\(V\) diagram, the process from \(P_1\) to \(P_2\) occurs at constant volume. Hence, \[ W = 0 \]
Step 2: Use first law of thermodynamics.
\[ Q = \Delta U + W \Rightarrow Q = \Delta U \]
Step 3: Write expression for change in internal energy.
\[ \Delta U = n C_v \Delta T \]
Step 4: Find temperature change using ideal gas equation.
At constant volume, \[ \frac{P}{T} = \text{constant} \] \[ \frac{T_2}{T_1} = \frac{P_2}{P_1} \Rightarrow \Delta T = T_1\left(\frac{P_2 - P_1}{P_1}\right) \]
Step 5: Substitute values.
\[ Q = n C_v \frac{(P_2 - P_1)V}{R} \] From graph, \(V = 1 \, \text{m}^3\).
\[ Q = \frac{10 \times 21 \times (30 - 21.7)}{8.3} = 21 \text{ J} \]
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