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Consider a cycle tyre being filled with ...

Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump `DeltaV(lt lt V)` of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from `P_1` to `P_2` ?

Text Solution

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The pressure is increased by `DeltaP`, when volume is increased by `DeltaV` at each stroke.
`therefore P_1V_1^gamma=P_2V_2^gamma` (Initially)
`therefore P(V+DeltaV)^gamma =(P+DeltaP)V^gamma` (`because` volume is fixed )
`PV^gamma [1+(DeltaV)/V]^gamma=P[1+(DeltaP)/P]V^gamma`
As `DeltaV lt lt V` so by using binomial theorem we get ,
`therefore PV^gamma (1+gamma (DeltaV)/V)=PV^gamma (1+(DeltaP)/P)`
`therefore gamma(DeltaV)/V=(DeltaP)/P`
`therefore DeltaV=1/gamma . V/P . DeltaP`
But `DeltaV` and `DeltaP` are very small.
`dV=1/gamma . V/P . dP`
Hence, work done in increasing the pressure in tube from `P_1` to `P_2` is
`W=int_(P_1)^(P_2)P dV=int_(P_1)^(P_2)Pxx1/gammaxxV/P.dP`
`=V/gamma int_(P_1)^(P_2)dP`
`=V/gamma (P_2-P_1)`
`therefore W=((P_2-P_1)V)/gamma`
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