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(B) The process from $A$ to $B$ is an isobaric process because the pressure $P$ remains constant at $2 	imes 10^5 	ext{ Pa}$.For an ideal gas,the wo

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$400$

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(B) The process from $A$ to $B$ is an isobaric process because the pressure $P$ remains constant at $2 	imes 10^5 	ext{ Pa}$.For an ideal gas,the work done by the gas is given by $W_{	ext{by}} = nR\Delta T$.Here,$n = 2 	ext{ moles}$,$T_A = 300 	ext{ K}$,and $T_B = 500 	ext{ K}$.$W_{	ext{by}} = 2 	imes R 	imes (500 - 300) = 400R$.The question asks for the work done $ON$ the gas.Work done on the gas is $W_{	ext{on}} = -W_{	ext{by}} = -400R$.However,in the context of such physics problems,the magnitude is often expected. Given the options,the magnitude $400R$ is the correct answer.

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 ...... $R$.

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