# Sign of work done by gas during isothermal expansion

$$\pu{10 mol}$$ of an ideal gas confined to a volume of $$\pu{10 L}$$ is released into atmosphere at $$\pu{300 K}$$ where the pressure is $$\pu{1 bar}.$$ The work done by the gas is:

Work done can be found as

$$W = -p_\mathrm{ext}(V_\mathrm{f} - V_\mathrm{i}).\tag{1}$$

$$V_\mathrm{f} = \frac{nRT}{p} = \frac{\pu{10 mol}\times\pu{0.083 L bar K^-1 mol^-1}\times\pu{300 K}}{\pu{1 bar}} = \pu{249 L}\tag{2},$$

$$W = -\pu{1 bar}\cdot(\pu{249 L} - \pu{10 L}) = \pu{-239 bar L}\tag{3}$$

As the question is asking us to find the work done by the gas, so shouldn't it be $$\pu{+239 bar L},$$ as work done by the gas is negative of the word done by external pressure?

But someone told me that the work done calculated is the work done by the gas and $$\pu{-239 bar L}$$ should be the answer. So, what mistake am I making and how do I determine if work done by the gas (and external pressure) is positive or negative under different conditions (expansion and compression)?

P.S.$$\pu{-239 bar L}$$ is the answer according to the answer key.

• It is the usual confusion. During invention of thermodynamics within physics, it was defined Delta U = Q - W, i.e. positive work is work done by the system. Chemistry later came with the opposite convention, Delta U = Q + W, with positive work done on system. Currently, physics is abandoning historical convention, accepting the one from chemistry. 6th Oct 18 '20 at 18:49
• Your answer for the work done by the gas is correct, and your friend and the answer key are wrong. Oct 18 '20 at 20:22

If we imagine our proverbial, frictionless piston containing some ideal gas, when $$P_{surr} > P_{syst}$$, the system is compressed by the piston and work is done on the system by the surroundings (i.e. work is positive). If the reverse is true ($$P_{syst} > P_{surr}$$), the gas in the piston will expand and work is done on the surroundings by the system (work is negative)