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JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

JEE Main 2023 Physics question on Thermodynamics

1 g of a liquid is converted to vapour at 3×1053 \times 10^5 Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm3^3 during this phase change, then the increase in internal energy in the process will be:
  1. A.
    4.32×1084.32 \times 10^8 J
  2. B.
    4800 J
  3. C.
    4320 J
  4. D.
    432000 J
Show answer and solution

Correct option: C

Solution

The work done in expansion is W = P·ΔV = 3×10⁵ Pa × 1600×10⁻⁶ m³ = 480 J.

This 480 J represents 10% of the total heat supplied, so total heat Q = 480/0.10 = 4800 J.

The increase in internal energy ΔU = Q − W = 4800 − 480 = 4320 J.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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