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 ltV) 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 ?

Consider a cycle tyre being filled with air by a pump. Let V be the vo

1.	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 ltV) 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 ?
Answer» Solution :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`