Thermodynamics - Result Question 34
36. A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are
$P_A=3 \times 10^{4} Pa$
$V_A=2 \times 10^{-3} m^{3}$
$P_B=8 \times 10^{4} Pa$
$V_D=5 \times 10^{-3} m^{3}$.
In process $A B, 600 J$ of heat is added to the system and in process $BC, 200 J$ of heat is added to the system. The change in internal energy of the system in process $A C$ would be
(a) $560 J$
(b) $800 J$
(c) $600 J$
(d) $640 J$
[1991]
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Answer:
Correct Answer: 36. (a)
Solution:
- (a) $B C$ is isobaric process,
$ \begin{aligned} & \therefore \quad W=P_B \times(V_D-V_A)=240 J \\ & \Delta Q=600+200=800 J \\ & \text{ Using } \Delta Q=\Delta U+\Delta W \\ & \Rightarrow \Delta U=\Delta Q-\Delta W=800-240=560 J \end{aligned} $
Since $A B$ is an isochoric process i.e., $\Delta V=0$ so, no work is done.
