One mole of an ideal gas undergoes a reversible process: T = T 0 + α V, where T 0 and α are constants. If its volume increases from V 1 to V 2 , the amount of heat transferred to the gas is

dT = α dV. Δ U = C V,m α(V 2 - V 1 ) = (C P,m - R)α(V 2 - V 1 ). W = ∫ P dV = ∫ (R(T 0 + α V))/VdV = RT 0 ln(V 2 )/(V 1 ) + Rα(V 2 - V 1 ). Q = Δ U + W = α C P,m (V 2 - V 1 ) + RT 0 ln(V 2 )/(V 1 ). Matches (c).

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ChemistryThermodynamics - IHeat & Heat CapacityEasy2 minQB

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One mole of an ideal gas undergoes a reversible process: T = T₀ + α V, where T₀ and α are constants. If its volume increases from V₁ to V₂, the amount of heat transferred to the gas is

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dT = α dV. Δ U = C_V,mα(V₂ - V₁) = (C_P,m - R)α(V₂ - V₁). W = ∫ P dV = ∫ (R(T₀ + α V))/VdV = RT₀ln(V₂)/(V₁) + Rα(V₂ - V₁). Q = Δ U + W = α C_P,m(V₂ - V₁) + RT₀ln(V₂)/(V₁). Matches (c).

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Heat & Heat Capacity

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