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Three processes compose a thermodynamics cycle shown in the $PV$ diagram. Process $1\rightarrow 2$ takes place at constant temperature. Process $2\rightarrow 3$ takes place at constant volume, and process $3\rightarrow 1$ is adiabatic. During the complete cycle, the total amount of work done is $10\,\, J$. During process $2\rightarrow 3$, the internal energy decrease by $20\,\,J$ and during process $3\rightarrow 1,$ $20\,\, J$ of work is done on the system. How much heat is added to the system during process $1\rightarrow 2\,\,?$ ...... $J$

$0$
$10$
$20$
$30$
Solution
Total work done in isothermal process is $=10+20=30 \mathrm{J}$
since in isothermal there is no change is internal energy, So all heat is converted to work. So the heat added to system is $30 \mathrm{J}$