Question

For cyclic process, the condition is:
A. $\Delta U = 0$
B. $\Delta H = 0$
C. $\Delta U > 0$ And $\Delta H > 0$
D. Both $\Delta U = 0$ and $\Delta H = 0$

Answer 1

Hint: In order to solve this question we need to understand cyclic process, enthalpy and internal energy. When the state of gas (real or ideal) changes from one point to another and then if we get the same state back then this process is known as cyclic process, the state change of gas is depicted by change in state parameters such as pressure, volume and temperature etc. Internal energy is defined as the sum of kinetic and potential energies of gases. According to the kinetic theory of gas, molecules of gas only have kinetic energy, so internal energy of gas is the total kinetic energy of molecules of gas.

Complete step by step answer:
Before answering let us understand state variables and path variables. State variables are those variables which do not depend on the path, it only depends on initial and final point.For example, Internal energy, Entropy, Enthalpy, Gibbs free energy and Helmholtz free energy

Path Variables are those variables which depend on the path by which the state of gas changes. For example, Work done, Heat etc. Since in a cyclic process, initial and final points are same and also we know that internal energy $U$ and Enthalpy $H$ are state variables, so there would be no change in internal energy and enthalpy as both initial and final points are the same in the cyclic process.
$\Delta U = 0\,\&\, \Delta H = 0$

So the correct option is D.

Note: It should be remembered that both the internal energy and enthalpy are thermodynamic potentials, so at equilibrium thermodynamic potential (Internal Energy, Enthalpy, Gibbs Free Energy and Helmholtz Free Energy) tends to minimize while entropy tends to maximize while entropy tends to maximize.