Question

The heat energy absorbed by a system in going through a cyclic process shown in figure is:

 

A. ${10}^7\pi J$
B. ${10}^6\pi J$
C. ${10}^2\pi J$
D. ${10}^4\pi J$

Answer 1

Hint: The graph of a cyclic process is given in the question. Here, the system starts the process and further returns to the same thermodynamic state. To find out the heat energy that is absorbed, we have to find out the work done in the entire process from the given P-V graph.

Formulas used:
$$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$$: First law of thermodynamics, where $$\mathrm{\Delta }Q$$ is the heat absorbed, $$\mathrm{\Delta }U$$ is the internal energy and $$\mathrm{\Delta }W$$ is the work done by the system.

Complete step by step answer:
Since the process is cyclic and it returns to the initial state, the change in internal energy doesn’t take place, i.e. $$\mathrm{\Delta }U=0$$
On applying the first law of thermodynamics, i.e. $$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$$,
We get the value of$$\mathrm{\Delta }Q=\mathrm{\Delta }W$$, as $$\mathrm{\Delta }U=0$$
Work done by the thermodynamic system is the enclosed area of the graph.
Hence, $$\mathrm{\Delta }Q=\mathrm{\Delta }W=\pi ab$$(area of the ellipse)
Here, a is the semi major axis and b is the semi minor axis of the ellipse.
$$a=b=10$$(Given)
Therefore $$\pi ab=$$$${10}^2\pi J$$=work done= total heat absorbed by the system.

Hence, option C is the right answer among the given options.

Note:Cyclic process is the underlying principle of the functioning of engines. When the graph goes clockwise, we can say that the value of W is positive and this represents a heat engine and if it goes counter clockwise, the value of W is negative and it will act as a heat pump.