# Work done by a gas in thermodynamics

1. A given mass of gas expands from state A to state B by three paths 1, 2, 3 as shown in the figure below.

If $$W_1,$$ $$W_2$$ and $$W_3$$ are respectively, be the work done by the gas along three paths, then

(A) $$W_1 > W_2 > W_3$$
(B) $$W_1 < W_2 < W_3$$
(C) $$W_1 = W_2 = W_3$$
(D) $$W_1 < W_2 : W_1 < W_3$$

The work done is $$W = -P \,\mathrm dV$$ that is the negative of the area under the graph; since Area 3 > Area 2 > Area 1, thus, $$W_1 > W_2 > W_3.$$

But the answer given is option (B). Why is this so?

• The question is a little confusing. Clearly the third path does the most work because the area under the curve is the largest. But if you compare them using the sign convention, it would look the smallest. I would say to always compared works by their magnitude rather than their sign, which really just tells us about direction. As a comparison if you wanted to know who was going faster, someone moving at -100 or -1 m/s, you wouldn't say the one moving at 1 m/s was faster. – Tyberius Apr 18 '17 at 14:15
• Can you say that since Work is a scalar we compare scalars by their magnitude? And by the way thank you for that beautiful analogy. – TheLostGuardian0 Apr 18 '17 at 14:27
• Yes, work is a scalar. From a math perspective this is because we can represent it as the dot product of two vectors (displacement and force). The sign is just for purposes of keeping track of whether the energy is moving to system or surroundings. – Tyberius Apr 18 '17 at 17:30

Sign conventions can make 'work' confusing, but the question here asks for the work done by the gas. When a gas expands, it's doing work (it's losing energy), so it's actually $W = + P \, \mathrm dV$, inverting all the inequality signs. So option B) is correct.

• But isn't it the standard convention in chemistry that work done by a gas means the system would lose energy hence -PdV? In physics it is +PdV. – TheLostGuardian0 Apr 18 '17 at 13:05
• The question asks for the work done by the gas. It's the only explanation I can think of, but maybe some other genius can shed more light on it. – Glorfindel Apr 18 '17 at 13:07
• @TheLostGuardian0, work adopts a sign convention. work done by and work done on are unambiguous, and do not. – a-cyclohexane-molecule Apr 18 '17 at 14:06
• In the same exam they had asked a previous question where they had given a P-V graph and asked the work done by the gas and the answer was -(area under graph). This was why my doubt arised. – TheLostGuardian0 Apr 18 '17 at 14:14
• The work done by the gas on the surroundings is the integral of +PdV. The work done by the surroundings on the gas is the integral of -PdV. So B is correct. – Chet Miller Apr 18 '17 at 14:23

Work done by the gas and on the gas are two different things.

In your question, the volume is increasing and the pressure is decreasing.

Let's consider some things before making an attempt to answer your question,

For simplification, consider the adiabatic process where $$Q=0$$.

If the external agent does positive work, the volume always decreases.

But in your question, as the volume is increasing, the external agent isn't the one doing the positive work, its actually the gas because only the gas can expand the system and still do positive work.

In the first law of thermodynamics,

$$\mathrm dU=Q+W$$

$$\mathrm dU=Q-p\,\mathrm dV$$,

The term $$W$$ represents the work done by the external agent.

If the external agent does positive work, volume decreases (i.e, $$\mathrm dV$$ is negative => $$W$$ is positive) and the internal energy increases and vice versa.

So, the work done by the gas becomes $$-(-p\,\mathrm dV)$$, because for a reversible process the work done by external agent and the gas are equal in magnitude and opposite in sign.

Now, as $$\mathrm dV$$ is positive, work done by the gas in also positive.

As you can see, magnitude wise $$W_3$$ is the greatest and now we also know that the net positive work is done by the gas because the final volume is greater.