The change in the magnitude of the volume of an ideal gas when a small additional pressure \(\Delta P\) is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity \(\Delta T\) at constant pressure. The initial temperature and pressure of the gas were \(300\, K\) and \(2\; atm\) respectively. If \(|\Delta T|=C|\Delta P|\) then value of \(C\) in \((K / a t m)\) is......

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Correct Answer

(A) \(150\)

Detailed explanation

\(PV = nRT\)

\(P \Delta V + V \Delta P =0 \quad(\) for constant temp. \()\)

\(P \Delta V = n R \Delta T \quad(\) for constant pressure)

\(\Delta T =\frac{ P \Delta V }{ nR }\)

\(\Delta P =-\frac{ P \Delta V }{ V } \quad(\Delta V\) is same in both cases \()\)

\(\frac{\Delta T }{\Delta P }=\frac{ P \Delta V }{ nR } \frac{ V }{- P \Delta V }=\frac{- V }{ nR }=-\frac{ T }{ P }\)

\(( PV = nRT )\)

\(\left(\frac{ V }{ nR }=\frac{ T }{ P }\right) \quad\left|\frac{\Delta T }{\Delta P }\right|=\left|\frac{-300}{2}\right|=150\)

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