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 AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be :

• A. 300 J
• B. 380 J
• C. 500 J
• D. 460 J

Answer 1

The correct option is D

460 J

See figure alongside

Process AB is isochoric so no work is done.

Heat added to be system is Q = 400 J.

$Q=\Delta U+\Delta W$

where $\Delta U$ is the change in internal energy $\Delta W$ is the work done.

Since $\Delta W=0$

$\Delta U=Q=400J$

Change in internal energy is 400 J.

Process BC is isobaric and the work done is given by

$\Delta W=P(V_2-V_1)=6\times {10}^4(4\times {10}^{-3}-2\times {10}^{-3}$

$=6\times {10}^4\times 2\times {10}^{-3}=120J$

Heat added to be system is Q = 100 J.

Since $Q=\Delta U+\Delta W$

$\therefore \Delta U=Q-\Delta W=(100-120)J=-20J$

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 AC is the area under the curve.

Area of the trapezium $=\frac{1}{2}(P_2+P_1)\times (V_2+V_1)$
$=\frac{1}{2}(6\times {10}^4+2\times {10}^4)\times (4\times {10}^{-3}-2\times {10}^{-3})$
$=\frac{1}{2}\times 8\times {10}^4\times 2\times 0^{-3}=80J.$

Since $Q=\Delta U+\Delta W$

and ∆U the change in internal energy in process AC, we have

$\Delta U=380Jand\Delta W=80J$

$\therefore Q=\Delta U+\Delta W=380+80=460J$