JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Heat & Thermodynamics

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

(A) $P_{2} \geq P_{1}$
(B) $P_{2} > P_{1}$
(C) $P_{1} = P_{2}$
(D) $P_{1} > P_{2}$ ✓ Correct

Answer: (D)

Solution

For an isobaric process,

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

.

On a

V

–

T

graph this is a straight line through the origin with slope

\frac{n R}{P}

, so the slope is inversely proportional to pressure — a smaller slope means a larger pressure.

In the given graph, line 1 has the smaller slope, so

P_{1} > P_{2}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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