For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows: (a) P1 > P2 (b) P1 < P2 (c) P1 = P2 (d) Insufficient data to draw any conclusion

Answer: (a). V = nR/PT, so at constant pressure V - T is a straight line of slope nR/P. Slope ∝ 1/P: the steeper line has the lower pressure. The P₂ line is steeper, so P₂ P₂.

For a perfect gas, two pressures P1 and P2 are shown in figure. The… | JEE Main 2022 (27 Jun, Shift 2)

Q55JEE Main 2022 · 27 JUN · SHIFT 2Medium

Gas Laws & Ideal Gas Equation

For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows:

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Answer: (a)

$V = \frac{nR}{P}T$ , so at constant pressure $V$ - $T$ is a straight line of slope $\frac{nR}{P}$ .
Slope $\propto \frac{1}{P}$ : the steeper line has the lower pressure.
The $P_2$ line is steeper, so $P_2 < P_1$ .
So the answer is (a) $P_1 > P_2$ .

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