Question

As a result of the isobaric heating by `DeltaT=72K`, one mole of a certain ideal gas obtain an amount of heat `Q=1.6kJ`. Find the work performed by the gas, the increment of its internal energy and `gamma`.

Answer 1

By the first law of thermodynamics
`Delta Q = Delta U + Delta W`
In an isobaric process
`Delta Q = C_(P)Delta T`
and `Delta U = C_(V)Delta T`
`:. C_(P) Delta T = C_(V) Delta T + Delta W`
or `Delta W = (C_(P) - C_(V)) Delta T`
or `Delta W = R Delta T` `( :. C_(P) - C_(V) = R)`
`:. Delta W = 8.3 xx 72 = 597.6 J`
`Delta Q = C_(P) Delta T`
`Delta Q = C_(P) Delta T`
`:. 1.6 xx 1000 = C_(P) xx 72`
`implies C_(P) = (1.6 xx 1000)/(72) = 22.2 J mol^(-1) K^(-1)`
`Delta U = Delta Q = Delta W = 1.6 xx 1000 - 597.6 = 1002.4 J`
But `Delta U = C_(V) Delta T`
`:. C_(V) = (1002.4)/(72) = 13.9 J mol^(-1) K^(-1)`
`:. gamma = (C_(P))/(C_(V)) = (22.2)/(13.9) = 1.60`