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Work Done in Reversible Adiabatic Process

One mole of a...

Question

One mole of a real gas is subjected to heating at constant volume from $(P_{1}, V_{1}, T_{1})$ state to $(P_{2}, V_{1}, T_{2})$ state. Then it is subjected to irreversible adiabatic compression against constant external pressure $P_{3}$ atm till system reaches final state $(P_{3}, V_{2}, T_{3})$ . If the constant volume molar heat capacity of real gas is $C_{v}$ . Find out correct expression for $Δ H$ from state 1 to state 3.

C_{v} (T_{3} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})

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C_{v} (T_{2} - T_{1}) + (P_{3} V_{2} - P_{1} V_{1})

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C_{v} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})

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C_{p} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})

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Solution

The correct option is C $C_{v} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$
$A (P_{1}, V_{1}, T_{1}) I s o c h o r i c h e a t i n g - --------- \to Δ H_{1} B (P_{2}, V_{1}, T_{2})$
$Δ H_{2} ↓ i r r . a d i a b a t i c p r o c e s s (q = 0)$
$Δ H_{o v e r a l l} = Δ H_{1} + Δ H_{2}$
$Δ H_{1} = Δ U_{1} + (P_{2} V_{1} - P_{1} V_{1})$
$Δ H_{1} = C_{V} (T_{2} - T_{1}) + (P_{2} V_{1} - P_{1} V_{1})$
$Δ H_{2} = Δ U_{2} + (P_{3} V_{2} - P_{2} V_{1})$
$Δ H_{2} = - P_{3} (V_{2} - V_{1}) + (P_{3} V_{2} - P_{2} V_{1})$
$Δ H_{2} = P_{3} V_{1} - P_{2} V_{1}$
$Δ H_{o v e r a l l} = C_{V} (T_{2} - T_{1}) + (P_{2} V_{1} - P_{1} V_{1}) + P_{3} V_{1} - P_{2} V_{1}$
$= C_{V} (T_{2} - T_{1}) + P_{3} V_{1} - P_{1} V_{1}$

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