I use the sign convention:
- Heat absorbed by the system = $q+$ (positive)
- Heat evolved by the system = $q-$ (negative)
- Work done on the system = $w +$ (positive)
- Work done by the system = $w -$ (negative)
Could anyone please tell me, that volume increased in system does positive or negative work?
Answer
Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.
Alternatively you could reason using the formula: $dU = dQ - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)
$dW = PdV.$
Therefore
$dU = dQ - PdV$.
Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.
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