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Isothermal Expansion of an Ideal Gas MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

Isothermal Reversible And Isothermal Irreversible is considered one the most difficult concept.
14 Questions around this concept.

Solve by difficulty

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy $\Delta U$ of the gas in joules will be

-500 J

-505 J

+505 J

1136.25 J

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Concepts Covered - 1

Isothermal Reversible And Isothermal Irreversible

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

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

$\\ {\text {When the volume of the gas changes from }} \\ {\mathrm{V}_{1}-\mathrm{V}_{2} \text { , the total work done (W) can be given as }} \\ {\mathrm{W}=\mathrm{P} . \int \cdot \mathrm{d} \mathrm{V}} \\ {\text { If we consider the external pressure (P) to be }} \\ {\text { constant than }} \\ {\mathrm{W}=\mathrm{P} \int \mathrm{d} \mathrm{V}=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)=\mathrm{P} \cdot \Delta \mathrm{V}} \\ {{\mathrm{W}=\mathrm{P} \cdot \Delta \mathrm{V}}}$

Isothermal irreversible expansion of an ideal gas

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

$\mathrm{W}=-\int_{\mathrm{V}_{1}}^{\mathrm{V}_{2}} \mathrm{P}_{\mathrm{ext}} \mathrm{dV}$

$\begin{array}{l}{\quad=-P_{e x t} \int_{V_{1}}^{V_{2}}} \\ {\quad =-P_{ex t}\left(V_{2}-V_{1}\right)} \\ {W=-P_{ext} \cdot \Delta V}\end{array}$

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

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

$\begin{array}{l}{\text { As according to ideal gas equation } \mathrm{PV}=\mathrm{nRT}} \\ {\quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}}\end{array}$

$\begin{array}{l}{\text { So } \mathrm{W}_{\mathrm{rev}}=\int \mathrm{\frac{nRT}{V}} \mathrm{dV} \text { (as temp. is constant) }} \\ {\left.\text { So } \mathrm{W}_{\mathrm{rev}}=-\mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}=-2.303 \mathrm{nRT} \log _{10} \frac{\left(\mathrm{V}_{2}\right)}{(\mathrm{V}_{1})}\right)} \\ {\mathrm{W}_{\mathrm{rev}}=-2.303 \mathrm{nRT} \log _{10} \frac{\left(\mathrm{P}_{1}\right)}{\mathrm{P}_{2}}}\end{array}$