The given figure represents two isobaric processes for the same mass of an ideal gas, then

For an ideal gas,

$$PV=nRT$$

For an isobaric process (P= constant),

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

This is a straight line on a V-T graph, with slope

$$slope=\frac{nR}{P}$$

Since same mass of gas is used, n is constant, so slope is inversely proportional to pressure:

$$\text{slope}\propto\frac{1}{P}$$

Line $$P_2$$ is steeper than line $$P_1$$, so

$$\frac{nR}{P_2}>\frac{nR}{P_1}$$

which gives

$$P_1>P_2$$

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