Question

When 1 mole of gas is heated at constant volume, the temperature rises from 273 K to 546 K. If heat supplied to the gas is \( x \, \text{J} \), then find the correct statement from following.

• A. \( Q = \Delta U = xJ, \, W = 0 \)
• B. \( Q = W = xJ, \, \Delta V = 0 \)
• C. \( \Delta V = 0, \, Q = W = -xJ \)
• D. \( Q = -W = xJ, \, \Delta V = 0 \)

Hint:

For any constant volume process, work done is always zero. The heat transfer equals the change in internal energy, regardless of the gas type.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have one mole of an ideal gas heated at constant volume. The temperature increases from 273 K to 546 K, and the heat absorbed is \( x \) joules. We must identify the correct thermodynamic relation among the options.

Step 2: Key Formula or Approach:
First law of thermodynamics: \( Q = \Delta U + W \). For a constant volume process, the work done \( W = \int P \, dV = 0 \) because \( dV = 0 \). Therefore, \( Q = \Delta U \).

Step 3: Detailed Explanation:
Since the volume does not change, no work is performed by or on the gas. All the heat supplied goes into increasing the internal energy of the gas. Hence, \( Q = \Delta U = x \, \text{J} \) and \( W = 0 \). The temperature rise confirms that internal energy increases.

Step 4: Final Answer:
Option (A) correctly states \( Q = \Delta U = xJ \) and \( W = 0 \).