Question

In an isothermal irreversible expansion of an ideal gas:
A. $\Delta U = 0$
B. $-Wq = nRT [ 1- \dfrac{P_{2}}{P_{1}}]$
C. $\Delta H \neq 0$
D. All of these

Answer 1

We are asked to find that what kind of change comes in an isothermal expansion of an ideal gas. We will first discuss the isothermal expansion of an ideal gas and then we will discuss the kind of change given in one of the options in an isothermal irreversible expansion of an ideal gas.

Complete step by step answer:
Isothermal irreversible expansion of an ideal gas : An isothermal process is a change in the system such that the temperature remains constant. In other words, in the isothermal process $\Delta T = 0$. Free expansion of gas occurs when it is subjected to expansion in a vacuum (Pex = 0). During the free expansion of an ideal gas, the work finished is zero be it a reversible or irreversible system.
Change in irreversible expansion of an ideal gas :
The internal energy of an ideal gas is directly proportional to the temperature of the gas. In this equation, R is the ideal gas constant in joules per mole kelvin and T is the temperature in Kelvin.
In the isothermal process, there may be no alternate in temperature so dT could be the same as 0. Hence, change in internal energy $\Delta U = 0.$ So the ideal solution is (A).
So, the correct answer is “Option A”.

Additional Information: Isothermal expansion in an ideal gas, all the collisions between molecules or atoms are perfectly elastic and no intermolecular force of attraction exists in an ideal gas because of the molecules of an ideal gas move so fast, and they are so far away from each other that they do not interact at all.

Note: Difference between isothermal process and free expansion : An isothermal process is a change in the system such that the temperature remains constant. In other words, in the isothermal process $\Delta T = 0.$