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 $5PV$ 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. $\frac{11}{2}PV$
• B. $6PV$
• C. $\frac{9}{2}PV$
• D. $7PV$

Answer 1

Answer: (A)

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

Work done, $W_1=p\Delta V=p(3V-V)=2pV$
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, $Q_1=5pV$ (given)
or $5pV=U_1+2pV$
Now for process $2$ ,
$\Rightarrow U_1=3pV$
Change of internal energy $=U_2=U_1=3pV$
As process $2$ and process $1$ lies between same isotherms.
Also, work done for process $2$ is
$W_2=$ Area under process $2$
$=\frac{1}{2}\times 2V\times \frac{1}{2}p+p\times 2V=\frac{5}{2}pV$
Again, using first law of thermodynamics, heat required in process $2$ is
$\Delta Q_2=\Delta U_2+\Delta W_2=3pV+\frac{5}{2}pV=\frac{11}{2}pV$