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

Under isobaric process `A = p Delta V = R Delta T (as v = 1)= 0.6 k J`
From the first law of thermodynamics
`Delta U = Q - A = Q - R Delta T = 1 k J`
Again increment in internal energy `Delta U = (R Delta T)/(gamma - 1), for v = 1`
Thus `Q - R Delta T = (R Delta T)/(gamma - 1)` or `gamma = (Q)/(Q - R Delta T) = 1.6`.