Question

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

Answer 1

Hint: 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 , $\mathrm{\Delta }V$ and hence $P\mathrm{\Delta }V$ is positive and so $\mathrm{\Delta }q$ 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 , $\mathrm{\Delta }V$ = change in volume
$P\mathrm{\Delta }V$ = the work done
$\mathrm{\Delta }q$ =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 .
$\Rightarrow \mathrm{\Delta }U=0$
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
$w=-P_{ext}\times \mathrm{\Delta }V$

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

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 .