Isothermal Expansion of an Ideal Gas - Practice Questions & MCQ

Isothermal reversible and irreversible 

Let us consider a cylinder fitted with a frictionless and weightless piston having an area of cross-section as 'A'. If the extemal pressure (P) is applied on this piston and the value of P is slightly less than that of the internal pressure of the gas When the gas undergoes a little expansion and the piston is pushed out by a small distance dx the work done by the gas on the piston is given by as

$\begin{aligned} & \mathrm{dw}=\text { force } \times \text { distance }=\text { pressure } \times \text { area } \times \text { distance } \\ & d w=P A \cdot d x \\ & \text { As A.dx }=\mathrm{dV} \\ & \mathrm{dw}=\mathrm{PdV}\end{aligned}$

When the volume of the gas changes from $\mathrm{V}_1-\mathrm{V}_2$, the total work done $(\mathrm{W})$ can be given as $\mathrm{W}=\mathrm{P} . \int \cdot \mathrm{dV}$
If we consider the external pressure $(\mathrm{P})$ to be constant than

$
\begin{aligned}
& \mathrm{W}=\mathrm{P} \int^{\mathrm{d}} \mathrm{~V}=\mathrm{P}\left(\mathrm{~V}_2-\mathrm{V}_1\right)=\mathrm{P} \cdot \Delta \mathrm{~V} \\
& \mathrm{~W}=\mathrm{P} \cdot \Delta \mathrm{~V}
\end{aligned}
$
 

Isothermal irreversible expansion of an ideal gas

When a gas expands against a constant external  $\left(P_{\text {ext }}=\right.$ constant $)$. There is a considerable difference between the gas pressure (inside the cylinder) and the external pressure. The temperature does not change during the process.

$\mathrm{W}=-\int_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{P}_{\mathrm{ext}} \mathrm{dV}$

$\begin{aligned} & =-P_{\text {ext }} \int_{V_1}^{V_2} \\ & =-P_{\text {ext }}\left(V_2-V_1\right) \\ W & =-P_{\text {ext }} \cdot \Delta V\end{aligned}$

Work done in Isothermal reversible expansion of an ideal gas 

As a small amount of work done dW on the reversible expansion of a gas through a small volume dV against an external pressure 'P' can be given as

$d W=-P d V$

So the total work done when the gas expands from initial volume V₁ to final volume V₂ is given as

$\int d W=\int_{v_1}^{v_2}-P d V$

As according to ideal gas equation $\mathrm{PV}=\mathrm{nRT}$

$
P=\frac{n R T}{V}
$
 

So $W_{\text {rev }}=\int \frac{n R T}{V} d V \quad$ (as temp. is constant)
So $\mathrm{W}_{\mathrm{rev}}=-\mathrm{nRT} \ln \frac{\mathrm{V}_2}{\mathrm{~V}_1}$

$
\begin{aligned}
& \mathrm{W}_{\mathrm{rev}}=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\
& \mathrm{~W}_{\mathrm{rev}}=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{P}_1}{\mathrm{P}_2}
\end{aligned}
$
 

Here negative sign indicates work of expansion and it is generally greater than work in the irreversible process

As in such a case, the temperature is kept constant and internal energy depends only on temperature so it internal energy is constant.

$\begin{aligned} & \text { So } \Delta E=0 \\ & \Delta E=q+W \\ & q=-W\end{aligned}$

Hence, during isothermal expansion, work is done by the system at the expense of heat absorbed.

Here $\Delta \mathrm{H}$ can be found out as follows:

$\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

As, for isothermal process, $\Delta E=0, \Delta T=0$ So $\Delta H=0$