Question

An ideal gas A and a real gas B have their volumes increased from $V$ to $2V$ under isothermal conditions. The increase in internal energy
A. will be same in both A and B
B. will be zero in both the gases
C. of B will be more than that of A
D. of A will be more than that of B

Answer 1

Hint:
This problem is based on the thermodynamic system of ideal and real gases. We know that the internal energy of real gases varies with the given conditions of the system and surroundings whereas the internal energy of ideal gases varies with the temperature only. Hence, use this approach to state the answer to the given problem.

Formula used:


Complete step by step solution:

By definition of an ideal gas, an ideal gas has no intermolecular forces and no particle-particle interaction. Therefore, a change in pressure at a steady temperature has no effect on internal energy.
Since internal energy is the function of temperature for an ideal gas.
 i.e., $U = f\left( T \right)$
Therefore, under isothermal conditions, $\Delta T = 0$ and thus, $\Delta {U_A} = 0$
However, the intermolecular forces are present in real gases, such as molecular attraction at low pressure and molecular repulsion at high pressure.
Since the volume is increased from $V$ to $2V$ , the effort must be made in opposition to these intermolecular forces i.e., work is to be done against real gas B which results in an increase in the Internal Energy of gas B.
$\Delta {U_B} \ne 0$
Thus, the increase in internal energy of B will be more as compared to A.
Hence, the correct option is (C) of B will be more than that of A.





Note:
In this problem, to determine the increase in the internal energy of an ideal gas A and a real gas B, we must find out that work is to be done against which gas during an increase in their volumes. Also, remember while writing an answer to this kind of conceptual problem, support your answer with exact reasons.