99 Percentile Qs Bank for JEE MainChemistryThermodynamics (C)
q ,   w ,   Δ E and Δ H for the following process A B C D for a monoatomic gas are:
Options
- Aw = - 2 P 0 V 0   ln   2 ,   q = 2 P 0 V 0   ln   2 ,   Δ E = 0 ,   &#
- Bw = - 2 P 0 V 0   ln   2 ,   q = 2 P 0 V 0   ln   2 ,   Δ E = 0 ,   &#
- Cw = - P 0 V 0 ( 1 + ln 2 ) ,   q = P 0 V 0 ( 1 + ln 2 ) ,   Δ E = 0 ,   Δ H = 0
- Dw = - P 0 V 0   ln   2 ,   q = P 0 V 0   ln   2 ,   Δ E = 0 ,   Δ
Correct answer
A. w = - 2 P 0 V 0   ln   2 ,   q = 2 P 0 V 0   ln   2 ,   Δ E = 0 ,   &#
Step-by-step solution
Temperature at A T A = P V 0 n R T B = 2 P 0 V 0 n R Since B C curve isothermal ∴ T C = 2 P 0 V 0 n R T D = P 0 / 2 · 2 V 0 n R = P 0 V 0 n R = T A Now ∵ T D - T A = 0 ∴ Δ U = Δ H = 0 ∵   For   ideal   gas   ΔU = nC V ΔT ΔH = nC P ΔT W T = W A B + W B C + W C D = - P 0 V 0 + - n R T ln 4 V 0 2 V 0 + - P 0 2 × - 2 V 0 W τ = - 2 P 0 V 0 ln 2 and q = - W T