Question

Figure below shows two paths that may be taken by a gas to go from a state $A$ to a state $C$.

In process $A B, 400\, J$ of heat is added to the system and in process $B C, 100 \, J$ of heat is added to the system. The heat absorbed by the system in the process $A C$ will be

• A. $300 \, J$
• B. $380 \, J$
• C. $500\, J$
• D. $460\, J$

Answer 1

Answer: (D)

According to given figure as shown below

Process $A B$ is isochoric,
i.e. $\Delta V=0$, so no work is done.
Heat added to the system in the process $A B$ is
$\Delta Q=400 \,J$
According to first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
where, $\Delta U$ is the change in internal energy and $\Delta W$ is the work done.
Since, $\Delta W=0$
$ \Rightarrow \Delta U=\Delta Q=400 \,J$
$\therefore$ Change in internal energy is $400 \,J$.
Process $B C$ is isobaric and the work done is given by
$\Delta W=p\left(V_{2}-V_{1}\right)$
$=6 \times 10^{4}\left(4 \times 10^{-3}-2 \times 10^{-3}\right)$
$=6 \times 10^{4} \times 2 \times 10^{-3}=120\, J$
Heat added to the system in the process $B C$ is
$\Delta Q=100\, J$
Since, $\Delta Q=\Delta U+\Delta W$
$\Delta U=\Delta Q-\Delta W=(100-120)\, J =-20\, J$
$\therefore$ Change in internal energy is $-20\, J$.
Total increase in internal energy is going from state $A$ to state $C$ is $400-20=380\, J$.
Work done in process $A C$ is the area under the curve. Area of the trapezium $A C D E A$
$=\frac{1}{2}\left(p_{2}+p_{1}\right) \times\left(V_{2}-V_{1}\right)$
$=\frac{1}{2}\left(6 \times 10^{4}+2 \times 10^{4}\right) \times\left(4 \times 10^{-3}-2 \times 10^{-3}\right)$
$=\frac{1}{2} \times 8 \times 10^{4} \times 2 \times 10^{-3}=80 \,J$
Again, by using $\Delta Q=\Delta U+\Delta W$,
where $\Delta U =380 \,J $ and $ \Delta W=80 \,J$
$\therefore \Delta Q =\Delta U+\Delta W$
$=380+80=460 \,J$
Thus, the heat absorbed by the system in the process $A C$ will be $460\, J$.