1 mole of an ideal gas expands isothermally reversibly from 2 litre to 4 litre and 3 moles of the same gas expands from 2 litre to x litre and doing same work, what is 'x'?

${(8)}^{1 / 3}$

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${(4)}^{2 / 3}$

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Solution

The correct option is B
${(4)}^{2 / 3}$

For reversible isothermal expansion,
$W = n R T l n \frac{V_{2}}{V_{1}}$
$W = n R T l n \frac{V_{2}}{V_{1}}$

Given,
$R T l n \frac{4}{2} = 3 R T l n \frac{x}{2}$

$l n 2 = l n {(\frac{x}{2})}^{3}$

$x^{3} = 16 \Rightarrow x = {(16)}^{\frac{1}{3}} = 4^{\frac{2}{3}}$
Hence, option (b) is correct.

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1 mole of an ideal gas expands isothermally reversibly from 2 litre to 4 litre and 3 moles of the same gas expands from 2 litre to x litre and doing same work, what is 'x'?

The correct option is B (4)2/3 For reversible isothermal expansion, W=nRTlnV2V1 W=nRTlnV2V1 Given, RTln42=3RTlnx2 ln 2=ln(x2)3 x3=16⇒x=(16)13=423 Hence, option (b) is correct.