Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram.
Assume the gas to be ideal the magnitude of work done on the gas in taking it form A to B is:

200 R

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300 R

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400 R

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500 R

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Solution

The correct option is C 400 R
Given : $T_{A} = 300 K$ $T_{B} = 500 K$ $n = 2$
Work done $W_{A B} = n R Δ T$ (isobaric process)
Or $W_{A B} = 2 R (500 - 300) = 400 R$

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Similar questions

D

A

A B C D A

P - T

R

	List - I		List - II
(P)	Magnitude of work done on the gas in taking from $A$ to $B$	(1)	693 R
(Q)	Magnitude of work done on the gas in taking from $B \to C$	(2)	277 R
(R)	Magnitude of work done on the gas in taking from $D \to A$	(3)	400 R
(S)	Magnitude of the net heat absorbed/evolved in the cycle $A B C D A$	(4)	416 R

A B C D A

A B C D A

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