JEE Main 2025 Apr 7 Shift 1, Physics Q25: Thermodynamic Processes

In a closed cycle on a P$-$V diagram, work done by the gas equals the enclosed area (with sign depending on the traversal direction).

The cycle is a circle (viewed in the P$-$V plane) with:

• Horizontal diameter (along P-axis): $\Delta P=500-300=200$ kPa $=200\times 10^3$ Pa
• Vertical diameter (along V-axis): $\Delta V=350-150=200$ cm${}^3=200\times 10^{-6}$ m${}^3$

Area of the ellipse-shaped loop (a circle in the axis-scaled diagram is generally an ellipse in true units; the area formula in physical units is):

$W=\frac{\pi }{4}\Delta P\cdot \Delta V$

Substitute:

$W=\frac{3.14}{4}\times (200\times 10^3)\times (200\times 10^{-6})$

$=\frac{3.14}{4}\times 40=\frac{125.6}{4}=31.4\text{ J}$

Expressing in units of $10^{-1}$ J: $W=314\times 10^{-1}$ J.

$\therefore$ Answer = 314.