Concept:The work done (W) by an expanding gas against an external pressure is given by \( P\Delta V \). If the volume is artificially kept constant, no physical expansion work can occur.
Formula:$$ W = P\Delta V $$
$$ q = \Delta E + P\Delta V $$
Solution:- At constant volume, the change in volume (\( \Delta V \)) is strictly zero.
- Therefore, the expansion work \( W = P \times 0 = 0 \).
- Substituting this back into the first law equation (\( q = \Delta E + W \)) yields: \( q_v = \Delta E + 0 \).
- The equation cleanly reduces to \( q_v = \Delta E \).
Why other options are incorrect:Option A and B still include the work term (\( P\Delta V \) or \( W \)), which violates the condition of constant volume. Option D lacks the equality to \( q_v \).
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