Question

Pick out the correct match from the following.

Options	Process	Feature
A.	Isothermal	Pressure constant
B.	Isobaric	Volume constant
C.	Isochoric	$\Delta Q = 0$
D.	Cyclic	$\Delta U = 0$

Answer 1

First, we have to identify the conditions for each of the given processes. Then, we will compare it with the features given corresponding to the process and identify whether it matches the process's conditions.

Complete answer:
First, let us take the first option A.As we know, an isothermal process is the physical change of a system where the temperature of the system maintains a constant value. An isothermal change includes isothermal expansion and isothermal compression. In isothermal expansion, the working substance of the system expands by absorbing heat keeping the temperature constant. In isothermal compression, the working substance inside the system releases heat and gets compressed, maintaining a constant temperature. Hence, we can say that the pressure is not constant in this process.Therefore, option A is incorrect.

Now, we consider option B.As we know, an isobaric process is the physical change of a system where the pressure has a constant value. In this process, the working substance of the system expands or contracts to maintain the pressure constant. The volume of the system changes in this process. Since the volume is not a constant in an isobaric process, we conclude that option B is incorrect.

Now, let us consider option C.In an isochoric process, the volume of a system will remain constant during a physical change of a system. Now, we express the first law of thermodynamics.
$\Delta U = \Delta Q - \Delta W\\\ \Rightarrow\Delta U = \Delta Q - P\Delta V$
where, $\Delta U$ is the change in internal energy of the system, $\Delta Q$ is the change in heat energy, $\Delta W$ is the work done, $P$ is the pressure and $\Delta V$ is the change in volume of the system.Since $\Delta V = 0$ for an isochoric process, $\Delta U = \Delta Q$. Also, $\Delta U \ne 0$. Hence, we obtain that $\Delta Q$ is not zero.Therefore, we find that option C is not correct.

Now, we consider option D.We know that in a cyclic process, the system comes back to its initial state after passing through many thermodynamic processes. Since the initial and the final states of the system are the same, we can say that the net change in the internal energy $\Delta U$ of the thermodynamic system will be zero. So, for a cyclic process, $\Delta U = 0$.

Therefore, option D is correct.

Note: In an isothermal process, the heat is exchanged with the surroundings slowly, ensuring that the temperature of the system does not change. Therefore, it is to be noted that in this process, the internal energy of the thermodynamic system does not change.Similarly, in every cyclic process, the internal energy of the system remains constant. But, in a non-cyclic process, the internal energy of the system changes.