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

$\begin{array}{rl}& \mathrm{dw}=\text{ force }\times \text{ distance }=\text{ pressure }\times \text{ area }\times \text{ distance }\\ & dw=PA\cdot dx\\ & \text{ As A.dx }=\mathrm{dV}\\ & \mathrm{dw}=\mathrm{PdV}\end{array}$

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

$\begin{array}{rl}& \mathrm{W}=\mathrm{P}{\int }^{\mathrm{d}}\mathrm{V}=\mathrm{P}({\mathrm{V}}_2-{\mathrm{V}}_1)=\mathrm{P}\cdot \mathrm{\Delta }\mathrm{V}\\ & \mathrm{W}=\mathrm{P}\cdot \mathrm{\Delta }\mathrm{V}\end{array}$
 

Isothermal irreversible expansion of an ideal gas

When a gas expands against a constant external  $(P_{\text{ext }}=$ 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.

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

$\begin{array}{rl}& =-P_{\text{ext }}{\int }_{V_1}^{V_2}\\ & =-P_{\text{ext }}(V_2-V_1)\\ W& =-P_{\text{ext }}\cdot \mathrm{\Delta }V\end{array}$

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

$dW=-PdV$

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

$\int dW={\int }_{v_1}^{v_2}-PdV$

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

$P=\frac{nRT}{V}$
 

So $W_{\text{rev }}=\int \frac{nRT}{V}dV$ (as temp. is constant)
So ${\mathrm{W}}_{\mathrm{rev}}=-\mathrm{nRT}\ln \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

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

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.

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

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

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

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

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