Two moles of an ideal gas are taken through the cyclic process abcda shown in a U–V diagram. During processes ab and cd, temperatures are 500 K and 300 K respectively. The volumes at a and d are V₀, and at b and c are 2V₀. Calculate the net heat absorbed by the system. (ln 2 = 0.69 and R = 8.3 J mol⁻¹ K⁻¹)
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Question

Two moles of an ideal gas are taken through the cyclic process abcda shown in a U–V diagram. During processes ab and cd, temperatures are 500 K and 300 K respectively. The volumes at a and d are V₀, and at b and c are 2V₀. Calculate the net heat absorbed by the system. (ln 2 = 0.69 and R = 8.3 J mol⁻¹ K⁻¹)

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G1 Physics · Accepted Answer

Step 1: Over a complete cycle, ΔU = 0 because internal energy is a state function. Step 2: Therefore the net heat absorbed equals the total work: Qₙₑₜ = Wₙₑₜ. Step 3: The horizontal upper branch a→b in the U–V graph is an isothermal expansion at 500 K, from V₀ to 2V₀. Step 4: Its work is W = nRT ln(Vfinal/Vinitial) = 2 × R × 500 × ln 2. Step 5: The lower branch c→d is isothermal compression at 300 K, from 2V₀ to V₀. Step 6: Its work is W = 2 × R × 300 × ln(1/2) = −2 × R × 300 × ln 2. Step 7: Th…