A certain volume of dry air at NTP is expanded reversible to four times its volume (a) isothermally (b) adiabatically. Calculate the final pressure and temperature in each case, assuming ideal behaviour.
(`(C_(P))/(C_(V))` for air=1.4)

Explanation

Let `V_(1)` be the intial volume of dry at NTP.
(a) Isotermal expansion: During isothermal expansion, the temperature remains the same throughout. Hence, final temperature will be 273K.
Since, `P_(1)V_(1)=P_(2)V_(2)`
`P_(2)=(P_(1)V_(1))/(V_(2))=(1xxV_(1))/(4V_(1))=0.25atm`
(b) Adiabatic expansion:
`(T_(1))/(T_(2))=((V_(2))/(V_(1)))^(gamma-1)`
`(273)/(T_(2))=((4V_(1))/(V_(1)))^(1.4-1)=4^(0.4)`
`T_(2)=(273)/(4^(0.4))=156.79K`
Final pressure: `(P_(1))/(P_(2))=((V_(2))/(V_(1)))^(gamma)`
`(1)/(P_(2))=((4V_(1))/(V_(1)))^(1.4)=4^(1.4)`
`P_(2)=(1)/(4^(1.4))=0.143`atm