JEE Advanced 2021 Paper 1 Q02 Physics Thermal Physics Kinetic Theory of Gases Medium

JEE Advanced 2021 Paper 1 · Q02 · Kinetic Theory of Gases

An ideal gas undergoes a four-step cycle on a $P$–$V$ diagram. In step 1, the gas expands isobarically (constant pressure) from a lower volume to a higher volume at high pressure. In step 2, the gas cools isochorically (constant volume) at the higher volume, so pressure decreases. In step 3, the gas is compressed isobarically at the lower pressure back to the original volume. In step 4, the gas is heated isochorically at the lower volume, so pressure increases back to the starting state. The cycle thus traces a rectangle on the $P$–$V$ diagram. During this cycle, heat is absorbed by the gas in

  1. A. steps 1 and 2
  2. B. steps 1 and 3
  3. C. steps 1 and 4
  4. D. steps 2 and 4
Reveal answer + step-by-step solution

Correct answer:C

Solution

Step 1 (isobaric expansion): ΔQ₁ = nC_p ΔT > 0 (heat absorbed). Step 2 (isochoric, P drops): ΔQ₂ = nC_v ΔT < 0 (heat released). Step 3 (isobaric compression): ΔQ₃ = nC_p ΔT < 0 (heat released). Step 4 (isochoric, P rises): ΔQ₄ = nC_v ΔT > 0 (heat absorbed). Therefore heat is absorbed in steps 1 and 4.

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