Question

A gas is expanded first isothermally and then adiabatically. In which case the work done will be greater?


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Answer 1

Given:

A gas is expanded first isothermally and then adiabatically.

To find:

In which case the work done will be greater?

Solution:

In these type of questions, its best to calculate the work done from P-V diagrams. We know that work is equal to the area under the P-V curve.

In Isothermal Expansion:

$PV = constant$

$\implies \: \dfrac{d(PV)}{d(V)} = 0$

$\implies \: V \dfrac{dP}{dV} + P= 0$

$\implies \: \dfrac{dP}{dV} = - \dfrac{P}{V}$

So, slope of the curve = -P/V.

In Adiabatic Expansion:

${PV}^{ \gamma } = constant$

$\implies \: \dfrac{d(P{V}^{ \gamma } )}{d(V)} = 0$

$\implies \: {V}^{\gamma} \dfrac{dP}{dV} + \gamma P {V}^{ (\gamma - 1)} = 0$

$\implies \: \dfrac{dP}{dV} = - \gamma \dfrac{P}{V}$

Since $\gamma$ > 1 ,

slope of adiabatic curve is greater than that of isothermal curve.

• So, area enclosed is higher in case of isothermal curve (in blue).

• Hence, work done is greater in Isothermal Expansion.