Question

In an isothermal irreversible expansion of an ideal gas :
A. $\Delta U = 0$
B. $- {w_q} = nRT[1 - \frac{{{P_2}}}{{{P_1}}}]$
C. $\Delta H \ne 0$
D. All of these

Answer 1

An isothermal process is one in which the pressure and volume of the system change but temperature remains constant . In case of isothermal irreversible expansion , the gas is allowed to expand slowly , its temperature tends to fall but some of the heat from the surrounding is conducted to the gas , keeping the temperature constant .

Complete step by step answer:
When a gas expands isothermally irreversibly , $\Delta V$ and hence $P\Delta V$ is positive and so $\Delta q$ will also be positive . Therefore , when a gas expands isothermally , an amount of heat equivalent to the work done by the gas has to be supplied from an external source .
Here , $\Delta V$ = change in volume
$P\Delta V$ = the work done
$\Delta q$ =heat
Now we know that in the isothermal process temperature remains constant . Therefore , since internal energy (U) is a function of temperature its value is also equal to zero .
$\Rightarrow \Delta U = 0$
In isothermal irreversible expansion , internal pressure is much different from external pressure . For example if internal pressure is much greater than the external pressure , then as the expansion takes place against the external pressure irreversibly , the work done is given by
$w = - {P_{ext}} \times \Delta V$

Hence option A is correct , that is $\Delta U = 0$ .

Note:
The work done in isothermal irreversible expansion is lesser than the work done in isothermal reversible expansion .
Also in the irreversible expansion , external pressure remains constant but in reversible expansion , external pressure has to be decreased continuously so as to remain infinitesimally smaller than the internal pressure .