Two moles of a monoatomic ideal gas go through a cycle ABCD with V A V B = 3 , V A V D = 6 and T A = 300 K. On a V - T graph AB is a straight line through the origin, BC has constant T ,

Question

Two moles of a monoatomic ideal gas go through a cycle ABCD with V A ​ V B ​ ​ = 3 , V A ​ V D ​ ​ = 6 and T A ​ = 300 K. On a V - T graph AB is a straight line through the origin, BC has constant T , CD has constant V and DA has constant T . Which statements are correct (heat in units of R )?

TestHub · Accepted Answer

AB isobaric: T B ​ = 3 T A ​ = 900 K and Q A B ​ = 2 × 2 5 ​ R × 600 = 3000 R . BC isothermal ( 3 V A ​ → 6 V A ​ ): Q = 2 R ( 900 ) ln 2 = 1247.7 R . CD isochoric ( 900 → 300 K): Q = 2 × 2 3 ​ R × ( − 600 ) = − 1800 R . DA isothermal ( 6 V A ​ → V A ​ ): Q = 2 R ( 300 ) ln 6 1 ​ = − 1075 R . Net work = Q n e t ​ = 1373 R , so D is wrong.

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