Which is the correct expression that relates changes of entropy with the change of pressure for an ideal gas at constant temperature, among the following?

Δ S = n R T ln \frac{P_{2}}{P_{1}}

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Δ S = T (P_{2} - P_{1})

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Δ S = n R ln (\frac{P_{1}}{P_{2}})

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Δ S = 2.303 n R T ln (\frac{P_{1}}{P_{2}})

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Solution

The correct option is C $Δ S = n R ln (\frac{P_{1}}{P_{2}})$
From first law of thermodynamics,
$Δ U = q + W$
$0 = q + W (for isothermal process)$
$q = - W = - (- P d V) = P d V$
$We know,$
$d S = \frac{q}{T} = \frac{P d V}{T} . . . (i)$
$By ideal gas equation,$
$P V = n R T$
$∴ P = \frac{n R T}{V} . . . (i i)$
Substituting equation $(i i)$ in equation $(i)$ ,
$d S = n R \frac{d V}{V}$
On integration from volume $V_{1}$ to volume $V_{2}$
$Δ S = n R ln (\frac{V_{2}}{V_{1}})$
Accordig to Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
Hence,
$Δ S = n R ln (\frac{P_{1}}{P_{2}})$ .

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