Question

In the process shown in figure, the internal energy of an ideal gas decreases by $\frac{3P_0{V}_0}{2}$ in going point C to A. Heat transfer along the process CA is

• A. $-3P_0{V}_0$
• B. $\frac{-5P_0{V}_0}{2}$
• C. $\frac{-3P_0{V}_0}{2}$
• D. $0$

Answer 1

The correct option is B $\frac{-5P_0{V}_0}{2}$

Work done is given by area under the PV graph.
Area under graph for CA process $=P_0(2V_0-V_0)=P_0{V}_0$
Since volume is decreasing, work done is $-P_0{V}_0$
Given that $\Delta U=-\frac{3P_0{V}_0}{2}$
The first law of thermodynamics gives us the relation:
$q=\Delta U+w$
$q=-\frac{3P_0{V}_0}{2}-P_0{V}_0=-\frac{5P_0{V}_0}{2}$
Hence option B is correct.