Question

In reversible isochoric change
A. $\Delta W = 0$
B. $\Delta Q = 0$
C. $\Delta T = 0$
D. $\Delta U = 0$

Answer 1

Hint: In this problem, to find out the correct option for a reversible isochoric change; we have to first find out the specific condition for the isochoric process and then analyze that condition with respect to different parameters such as work done, internal energy, temperature, etc., to give a correct result.

Complete answer:
Isochoric process in thermodynamics is a process during which the volume of a system remains constant that’s why it is also referred to as a constant-volume process.

A reversible process in thermodynamics is a process in which both the system and the surroundings return to their original state without any external effect on the universe.
According to the question, An Isochoric change to be reversible, the system must be maintained in thermal equilibrium and heat must be supplied in a quasi-static manner.

Also, we know that Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure, $Workdone = W = P\Delta V$ … (1)

Graphically, an isochoric process can be represented as: -

Also, By definition of an isochoric process, $\Delta V = 0$

So, from eq. (1), we get
$ \Rightarrow W = P\Delta V = 0$

Thus, In a reversible isochoric change $\Delta W = 0$. Hence, the correct option is (A) $\Delta W = 0$

Note: It is crucial to carefully consider all the given possibilities i.e., work done, internal energy, temperature, etc., in order to provide a comprehensive explanation because this is a theoretical conceptual problem. Always keep in mind to give specific justifications with the help of graphs in support of your explanation while writing an answer for this type of conceptual problem.