# Work done by a gas

An ideal gas changes from state $a$ to state $b$ as shown in the figure. What is the work done by the gas in the process?

This is how I reasoned it:

As the pressure in the gas increases, its temperature also increases. This the gas was not allowed to exchange heat with the surrounding so we can say it’s an adiabatic process. Now the work done on the gas would be negative (since temperature increased in increasing the pressure so we can assume that gas was compressed) and the work done by the gas would be zero (since the system was not allowed to exchange heat with the surroundings). Is my reasoning correct? What would be a more mathematical way to solve the problem?

• The work done on the gas and the work done by the gas is the same thing, only with different sign. – Ivan Neretin Dec 25 '17 at 7:09
• @Ivan Oh right! So where did I go wrong ? – Aditi Dec 25 '17 at 7:23
• Who said the work done on the gas is negative? True, the pressure increased, so what? It seems to have increased thanks to temperature alone. There was no work. – Ivan Neretin Dec 25 '17 at 7:55
• @Ivan sorry , my bad ,I thought the PV diagrams were drawn for constant temperature – Aditi Dec 25 '17 at 9:20
• Maybe (though then again, maybe not), but this is not PV. – Ivan Neretin Dec 25 '17 at 9:29

Your reasoning seems correct, but you shouldn't assume the nature of processes unless stated. I find that very uncertain, and unnecessary as in your question.

From the given diagram it is clear that there is a liner relation between $T$ and $P$,

$T=kP$ $\space$(Identical to the standard equation of a straight line, $y=mx$)

The questions says that the gas is ideal, which enables you to use ideal gas equation.

$PV=nRT$,

$V=\frac{T}{P}nR$

$V=knR=constant$

Now according to you, when the temperature increases, the pressure increases but the work is zero. The flaw is this reasoning could could be understood by considering the ideal gas equation. This reasoning only holds when the change in $T$ with respect to $P$ is constant (which is the definition of the slope of an line, $k$) which leaves you with a volume, but there can be cases where the graph between $T$ and $P$ is not a straight line, and in those cases work by the gas won't be zero.

As @Ivan added in his comment, you second statement which states that the work is zero, contradicts with your first one. This is not a very clear way to look at the process indicated by the diagram.

In case of adiabatic process, it is not necessary that work will be nil if no heat exchange takes place. It just means that heat exchange is zero, it does not tell you anything about the work done by (or on) the system.

What you need here is a change of perspective, you are thinking the temperature increases when the pressure increases (which indicates compression), but instead, you can increase temperature at constant volume which will increase the system's pressure.