Question

The three processes in a thermodynamic cycle shown in the figure are :
Process $1\to 2$ is isothermal;
Process $2\to 3$ is isochoric (volume remains constant);
Process $3\to 1$ is adiabatic

The total work done by the ideal gas in this cycle is $10J.$ The in ternal energy decreases by $20J$ in the isochoric process. The work done by the gas in the adiabatic process is $-20J.$ The heat added to the system in the isothermal process is

• A. 0 J
• B. 10 J
• C. 20 J
• D. 30 J

Answer 1

Answer: (D)

Work done in complete cycle
$W=W_{12}+W_{23}+W_{31}....(i)$

Given $W=10J,W_{23}=0$
and $W_{31}=-20J$
So, from Eq. (i), we have $W_{12}=30J$
As in isothermal process,
Heat absorbed = Work done
$\therefore$ Heat absorbed in process $1\to 2=30J$