Two moles of Helium gas are taken over the cycle ABCDA as shown in the P-T diagram. Assuming the gas to be ideal the work done on the gas in taking it from $A$ 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 $400$ $R$
Work done in isobaric process, $W_{A B} = P Δ V = n R (T_{f} - T_{i}) = 2 \times R (500 - 300) = 400 R$

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Similar questions

Q. Two moles of helium gas are taken over the cycle

A B C D A

, as shown (below) in the

P - T

diagram. (Assume the gas to be ideal and

R

is gas constant).
Now, match List I with List II and select the options given below:

	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

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A B C D A

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Two moles of Helium gas are taken over the cycle ABCDA as shown in the P-T diagram. Assuming the gas to be ideal the work done on the gas in taking it from A to B is :

The correct option is C 400 RWork done in isobaric process,WAB=PΔV=nR(Tf−Ti)=2×R(500−300)=400R

Two moles of Helium gas are taken over the cycle ABCDA as shown in the P-T diagram. Assuming the gas to be ideal the work done on the gas in taking it from $A$ to $B$ is :

The correct option is C $400$ $R$
Work done in isobaric process, $W_{A B} = P Δ V = n R (T_{f} - T_{i}) = 2 \times R (500 - 300) = 400 R$