An ideal gas undergoes a process maintaining relation between pressure (P) and Volume (V) as \(P={P}_{0}{\left(1+{\left(…
An ideal gas undergoes a process maintaining relation between pressure (P) and Volume (V) as \(P={P}_{0}{\left(1+{\left(\frac{{V}_{0}}{V}\right)}^{2}\right)}^{-1}\) where \({P}_{0}\) and \({V}_{0}\) are constants. If two samples A and B (two moles each) with initial volumes \({V}_{0}\) and \(3{V}_{0}\) respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples \({T}_{B}-{T}_{A}\)_____.(R= gas constant)
[04 April, 2026 (Shift-I)]
\(\frac{11{P}_{0}{V}_{0}}{10R}\)
At V=\({V}_{0}\), the pressure is:
\({P}_{A}=\frac{{P}_{o}}{1+{(\frac{{V}_{o}}{{V}_{o}})}^{2}}=\frac{{P}_{o}}{2}\)
At \(V=3{V}_{0}\) the pressure is
\({P}_{B}=\frac{{P}_{o}}{1+{(\frac{{V}_{o}}{3{V}_{o}})}^{2}}=\frac{{P}_{o}}{1+\frac{1}{9}}=\frac{9{P}_{o}}{10}\)
Using \(T=\frac{PV}{nR}\) with n=2:
For sample A (V= \({V}_{0}\))
\({T}_{A}=\frac{{P}_{0}{V}_{0}}{4R}\)
For sample B (V=\(3{V}_{0}):\)
\({T}_{B}=\frac{{P}_{B}(3{V}_{o})}{2R}=\frac{(\frac{9{P}_{o}}{10})(3{V}_{o})}{2R}=\frac{27{P}_{o}{V}_{o}}{20R}\)
Temperature Difference
\({T}_{B}−{T}_{A}=\frac{27{P}_{o}{V}_{o}}{20R}−\frac{{P}_{o}{V}_{o}}{4R}\\ {T}_{B}−{T}_{A}=\frac{11{P}_{o}{V}_{o}}{10R}\)
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