Question

Three processes form a thermodynamic cycle as shown on P - V diagram for an ideal gas.
Process $1\to 2$ takes place at constant temperature $300\text{K}$.
Process $2\to 3$ takes place at constant volume.
During this process $40\text{J}$ of heat leaves the system.
Process $3\to 1$ is adiabatic and temperature $T_3$ is $275\text{K}$.
Work done by the gas during the process $3\to 1$ is

• A. $-40\text{J}$
• B. $-20\text{J}$
• C. $+40\text{J}$
• D. $+20\text{J}$

Answer 1

The correct option is A $-40\text{J}$


For process $2\to 3-$ volume constant
$Q=-40\text{J}$
$W=0$ (As $V$ is constant)
$\Delta U=Q=-40$
$U_3-U_2=-40.....\left(i\right)$
For process $1\to 2-$ temperature constant
$T_1=T_2$
$U_1=U_2....\left(ii\right)$

From $\left(i\right)\&\left(ii\right)$
$U_3-U_1=-40$
$\Delta U_{3\to 1}=40\text{J}$

For process $3\to 1-$ Adiabatic process
$W=-\Delta U$
$W=-40\text{J}$