# Thermodynamics - Sign convention

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?

## 3 Answers

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.

I think the confusion arises due to difference in equations from Physics and Chemistry text books.

The physics text writes ∆U=q-w while chemistry says ∆U=q+w.
From the physics text its mentioned that work done by the system is positive means ∆U=q-(+w) and work done on the system is negative means ∆U=q-(-w).

THEREFORE, both are correct in their own way. Hope I've figured it out correctly.

to solve this sign conflict between physics and chemistry, you may, in the two cases, only refer to the decrease of the internal energy, U, as an indication of work done by the system, and refer to the increase of the internal energy, U, as an indication of work done on the system.