Question:

When the expansion of a gas occurs in vacuum and at constant volume, then:

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In an expansion at constant volume and in a vacuum, no work is done. Hence, the change in internal energy is equal to the heat added to the system.
Updated On: Apr 23, 2025
  • \( \Delta U = qV \)
  • \( \Delta U = qp \)
  • \( \Delta H = qV \)
  • \( \Delta H = qp \)
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The Correct Option is A

Solution and Explanation


In thermodynamics, the first law is given by: \[ \Delta U = Q - W \] Where:
- \( \Delta U \) is the change in internal energy,
- \( Q \) is the heat added to the system,
- \( W \) is the work done by the system.
For a gas expansion in a vacuum, the work done \( W \) is zero because there is no external force against which the gas does work. Therefore, the first law of thermodynamics simplifies to: \[ \Delta U = Q \] Now, considering that the gas is expanding at constant volume, the heat added to the system \( Q \) is proportional to the change in internal energy, and we use \( Q = qV \), where \( qV \) is the heat absorbed at constant volume. Thus, the change in internal energy is: \[ \Delta U = qV \] Therefore, the correct answer is option (A).
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