1. Use the ideal gas law for an isobaric process

For a fixed mass of an ideal gas, $PV=nRT$ If the process is isobaric, then $P$ is constant, so $V=\frac{nR}{P}T$ This shows that on a $V$ vs $T$ graph, an isobaric process is a straight line through the origin with slope $m=\frac{V}{T}=\frac{nR}{P}$

So, for the same gas mass:

• larger slope $\Rightarrow$ smaller pressure,
• smaller slope $\Rightarrow$ larger pressure.

2. Compare the two isobaric lines in the figure

From the figure, the line corresponding to $P_1$ is steeper than the line corresponding to $P_2$.

Hence, $\frac{nR}{P_1}>\frac{nR}{P_2}$ Cancelling the common positive factor $nR$, $\frac{1}{P_1}>\frac{1}{P_2}$ Therefore, $P_1<P_2$

3. Choose the correct option

Thus, $P_2>P_1$ So the correct option is:

A: $P_2>P_1$

4. Compare with stored correct answer

Stored correct answer: B: $P_1>P_2$

Our derived result is A: $P_2>P_1$. Therefore, the stored answer does not agree with the physics of isobaric lines on a $V$-$T$ graph, assuming the steeper line is labeled $P_1$ and the less steep line is labeled $P_2$.