NEET-XII-Physics
26: Laws of Thermodynamics
- #6An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas?Ans : Let:
P1 = Initial pressure
P2 = Final pressure
T1 = Absolute initial temperature
T2 = Absolute final temperature
Given :
`` \frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}``
Using the ideal gas equation, we get
PV = nRT
If n is the number of moles of the gas and R is the universal gas constant, then
`` \frac{{P}_{1}}{{T}_{1}}\mathit{=}\frac{\mathit{n}\mathit{R}}{{\mathit{V}}_{\mathit{1}}}\,\mathrm{\,and\,}\frac{{P}_{2}}{{T}_{2}}\mathit{=}\frac{\mathit{n}\mathit{R}}{{\mathit{V}}_{\mathit{2}}}``
`` \Rightarrow {V}_{1}\mathit{=}{V}_{2}\left[\because \frac{{P}_{\mathit{1}}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}\right]``
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`` \mathit{\Rightarrow }\mathit{∆}V={V}_{2}\mathit{-}{V}_{1}=0``
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Thus, Work done by gas = P∆V = 0
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