Question

The $P-V$ diagram shown below indicates two paths along which a sample of gas can be taken from state $A$ to state $B$. The energy equal to $5 PV$ in the form of heat is required to be transferred if the Path $-1$ is chosen. How much energy in the form of heat should be transferred if Path-$2$ is chosen?

• A. $\frac{11}{2} P V$
• B. $6 \,P V$
• C. $\frac{9}{2} P V$
• D. $7 \,P V$

Answer 1

Answer: (A)

Work done in a thermodynamic process
$=$ Area under $p-V$ graph
For process $1$,

Work done, $W_{1}=p \Delta V=p(3 V-V)=2 p V$
Let change of internal energy in process $1$ (point $A$ to point $B$ ) is $U_{1}$.
Then, by first law of thermodynamics, heat required is
$\Delta Q_{1}=\Delta U+\Delta W=U_{1}+W_{1}$
Here, $Q_{1}=5 p V$ (given)
or $5 p V=U_{1}+2 p V$
Now for process $2$ ,
$\Rightarrow U_{1}=3 p V$
Change of internal energy $=U_{2}=U_{1}=3 p V$
As process $2$ and process $1$ lies between same isotherms.
Also, work done for process $2$ is
$W_{2}=$ Area under process $2$
$=\frac{1}{2} \times 2 V \times \frac{1}{2} p+p \times 2 V =\frac{5}{2} p V$
Again, using first law of thermodynamics, heat required in process $2$ is
$\Delta Q_{2}=\Delta U_{2}+\Delta W_{2}=3 p V+\frac{5}{2} p V=\frac{11}{2} p V$