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Isothermal Expansion of an Ideal Gas - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • Isothermal Reversible And Isothermal Irreversible is considered one the most difficult concept.

  • 16 Questions around this concept.

Solve by difficulty

What will be the magnitude of work done when at \mathrm{STR,\ 3.2 \mathrm{~g}} of oxygen is expanded to occupy for times its original volume?

What will be \mathrm{\Delta E} if at \mathrm{500 K}, I mole of an ideal gas was isothermally expanded from \mathrm{1L} to \mathrm{2L} ?

A gas sample occupies \mathrm{ 2.5 \mathrm{~L}} at \mathrm{ 27^{\circ} \mathrm{C}} and 1 bar pressure. If the gas undergoes an isothermal expansion to \mathrm{ 10 \mathrm{~L}}, what is the final pressure?

A gas of amount 0.2 mole undergoes a reversible isothermal compression at a constant temperature of 300 \mathrm{~K}. The gas is compressed from an initial volume of 10 liters to a final volume of 5 liters. Calculate the work done on the gas during the process.

A sample of one mole of a diatomic ideal gas initially occupies a volume of 8 liters at a temperature of 400 \mathrm{~K}. The gas is allowed to expand isothermally to a final volume of 20 liters. Calculate:
 

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} \quad \quad \text { (as temp. is constant) }} \\ \\ \text {So } \mathrm{W}_{\mathrm{rev}}= - \mathrm{nRT} \ln \frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}\\\\ \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}{ll}{\text { So } \ \Delta E} {\ =0} \\ {\Delta E=q+W} \\ {q=} {-W}\end{array}

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

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

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

\begin{array}{l}{\text { As, for isothermal process, } \Delta E=0, \Delta T=0} \\ {\text { So } \Delta H=0}\end{array}

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Isothermal Reversible And Isothermal Irreversible

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Reference Books

Isothermal Reversible And Isothermal Irreversible

Chemistry Part I Textbook for Class XI

Page No. : 168

Line : 32

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