If `1 gram` of oxygen at `760 mm` pressure and `0^(@)C` has its volume double in an adiabatic change , calculate the change in internal energy. Take `R=2 cal. Mol e^(-1)K^(-1)= 4.2J cal^(-1) and gamma=1.4`.

Answer

`-43.37J`

Explanation

In an adiabatic change, `dQ=0`
From `dU+dW=dQ=0`
`dU= -dW= -(r(T_(2)-T_(1)))/(1-gamma)`
For `T_(2), use T_(2)=T_(1)((V_(1))/(V_(2)))^((gamma-1))` and `r=R/M`.