- Home
- Standard 11
- Physics
Three processes form a thermodynamic cycle as shown on $P-V$ diagram for an ideal gas. Process $1 \rightarrow 2$ takes place at constant temperature $(300K$). Process $2 \rightarrow 3$ takes place at constant volume. During this process $40J$ of heat leaves the system. Process $3 \rightarrow 1$ is adiabatic and temperature $T_3$ is $275K$. Work done by the gas during the process $3 \rightarrow 1$ is ..... $J$
$-40$
$-20$
$+40$
$+20$
Solution
In the process $1 \rightarrow 2, \Delta Q_{1 \rightarrow 2}=\Delta W_{1 \rightarrow 2}$
In the process $2 \rightarrow 3, \Delta W_{2 \rightarrow 3}=0$
In the process $3 \rightarrow 1, \Delta Q_{3 \rightarrow 1}=0$
The complete process being cyclic, $\Delta U=0$
Hence $\Delta Q=\Delta W \Rightarrow \Delta Q_{1 \rightarrow 2}+\Delta Q_{2 \rightarrow 3}+\Delta Q_{3 \rightarrow 1}=\Delta W_{1 \rightarrow 2}+\Delta W_{2 \rightarrow 3}+\Delta W_{3 \rightarrow 1} \Rightarrow$
$\Delta Q_{2 \rightarrow 3}=\Delta W_{3 \rightarrow 1}=-40 J$
