The given figure | KTG & THERMODYNAMICS |JEE-MAIN 2023

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JEE MAIINS 2024

31.01.24

Question

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

Select the correct option:

$P_2 \geq P_1$

$P_2>P_1$

$P_1=P_2$

$P_1>P_2$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Ans. (4) Sol. $$ \begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{~V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T} \\ & \text { Slope }=\frac{\mathrm{nR}}{\mathrm{p}} \\ & \text { Slope } \propto \frac{1}{\mathrm{p}} \\ & (\text { Slope }))_2>\text { (Slope) } \\ & \mathrm{P}_2<\mathrm{P}_1 \end{aligned} $$

Question Tags

JEE Main

Physics

Medium

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