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  • Three moles of an ideal gas are taken through a cyclic process ABCA as shown on T − V diagram in Fig. The gas loses 2510 J of heat in the complete cycle. If <img alt="\mathrm{T_A=100 \

Three moles of an ideal gas are taken through a cyclic process ABCA as shown on T − V diagram in Fig.
The gas loses 2510 J of heat in the complete cycle. If \mathrm{T_A=100 \mathrm{~K} \text { and } T_B=200 \mathrm{~K}}, The work done by the gas
during the process BC is

Option: 1

5000 J


Option: 2

−5000 J


Option: 3

4000 J


Option: 4

−2500 J


Answers (1)

best_answer

In process AB, the volume V increases linearly with temperature T. Hence process AB is
isobaric (constant pressure). Therefore, work done in this process is

\mathrm{\begin{aligned} & \quad W_{A B}=P \Delta V=n R \Delta T(\because P V=n R T) \\ & \left.=n R\left(T_B-T_A\right) \quad 3 \times 100\right)=2490 \mathrm{~J} \\ & =3 \times 8.3 \times(200-100) \end{aligned}}

Process CA is isochoric (constant volume). Hence work done in this process \mathrm{W_{C A}=0} Since the whole process ABCA is cyclic, the change in internal energy in the complete cycle is zero, i.e. \mathrm{\Delta U=0}. Now, from the first law of thermodynamics, (Given Q =
−2510J)

\mathrm{\begin{aligned} & Q=\Delta U+W=\Delta U+W_{A B}+W_{B C}+W_{C A} \\ & \text { Or }-2510=0+2490+W_{B C}+0 \\ & \text { Or } W_{B C}=-2510-2490=-5000 \mathrm{~J} \end{aligned}}

The negative sign shows that the work is done by the gas

Posted by

himanshu.meshram

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