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An ideal gas goes through the cycle abc. For the complete cycle 800J of heat flows out of the gas. Process ab is at constant pressure and process bc is at constant volume. In process c-a, `ppropV`. States a and b have temperature `T_a=200K` and `T_b=300K`. (a) Sketch the p-V diagram for the cycle. (b) What is the work done by the gas for the process ca?

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The correct Answer is:
B, D

(a)
`W_(n et)=Q_(n et)` in a cycle.
Since, `Q_(n et)` is negative (flows out of the gas). Hence, `W_(n et)` should aslo be negative. Or, cycle should be anti-clockwise as shown in figure.
From a to b
p=constant
`:.` `VpropT`
T has become 1.5 times. Therefore, V will also become 1.5 times.
From c to a
`p propV`
V has become `1/1.5` times. Therefore,
p will become `1/1.5` times.
In a cycle
`|Q_(n et)|=|W_(n et)|`=Area of cycle
`:.` `800=1/2(0.5p_0)(0.5V_0)`
`:.` `p_0V_0=6400J`
`W_(ca)=-Area under the graph
`=-1/2(2.5p_0)(0.5V_0)`
`=-0.625p_0V_0`
`=-0.625xx6400=-4000J`
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