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Figure Shows a process ABCA performed on...

Figure Shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process.

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As the process is cycle, the change in internal energy is zero. The heat given to the system is then equal to the work done byh it.
The work done in part `AB is W_(1)=0` as the volume remains constant. The part BC represent an isothermal procces so that the work done by the gas during this part is
`W_(2)=nRT_(2)In(V_(2)//V_(1))`
during the part CA,
`VpropT`.
so, `V//T` is constant and hence,
`p=(nRT)/V` is constant.
The work done by the gas during the part CA is ` the work done by the gas the part CA is
`W_(3)=p(V_(1)-V_(2))`
`=nRT_(1)-nRT_(2)`
`=-nR(T_(2)-T-(1).
The net work done by the gas in the process ABCA is
W=W_(1)+W_(2)+W_(3)=nR[T_(2)InV_(2)/V_(1)-(T_(2)-T_(1))]`.
The same amount of heat is given to the gas.
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