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

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

$P_{2} \geq P_{1}$
$P_{2} > P_{1}$
$P_{1} = P_{2}$
$P_{1} > P_{2}$

Solution

As we know, $P V = n R T$

$\Rightarrow V = (\frac{n R}{P}) T$

Therefore, slope $= \frac{n R}{P}$

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

As ${(Slope)}_{2} > {(Slope)}_{1}$

$\Rightarrow P_{2} < P_{1}$

Asked in: JEE Main 2024 (31 Jan Shift 1)

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