An ideal monoatomic gas undergoes a process where its pressure is inversely proportional to its temperature. If the work done by two moles of gas if the temperature changes from $T_{1} t o T_{2}$ is $x R (T_{2} - T_{1})$ . Find $x$

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Solution

Given that pressure is inversely proportional to temperature:
$P \propto \frac{1}{T}$
i.e.
$P = \frac{k}{T}$
Ideal gas law states that:
$P V = n R T$
$T = \frac{P V}{n R}$
i.e
$P = \sqrt{\frac{n R k}{V}}$
Work done $= \int P d V = \int \sqrt{\frac{n R k}{V}} d V$
Work done $= 2 \sqrt{n R k} (\sqrt{V_{2}} - \sqrt{V_{1}})$
Also from the previous relations:
$(\sqrt{\frac{n R}{k}}) T = \sqrt{V}$ Substituting this in 'Work done'
Work done $= 2 n R (T_{2} - T_{1})$
For two moles of gas.
Work done $= 4 R (T_{2} - T_{1})$
Hence $x = 4.$

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An ideal monoatomic gas undergoes a process where its pressure is inversely proportional to its temperature. If the work done by two moles of gas if the temperature changes from T1toT2 is xR(T2−T1). Find x

An ideal monoatomic gas undergoes a process where its pressure is inversely proportional to its temperature. If the work done by two moles of gas if the temperature changes from $T_{1} t o T_{2}$ is $x R (T_{2} - T_{1})$ . Find $x$