Question

Fixed mass of an ideal gas contained in `10.0L` sealed rigid vessel at `1atm` is heated from `-73^(@)C` to `27^(@)C`. Calculate change in Gibbs enegry if entropy of gas is a function of temperature as `S = 2 +10^(-2)T (J K^(-1)). (1atmL = 0.1kJ)`

Answer 1

Using: `dG = V dP - S dT`
`rArr DeltaG = int V dP - int S dT = VDeltaP -int_(T_(1))^(T_(2)) (2+10^(-2)T)dT`
`=V(P_(2)-P_(1)) -2(T_(2)-T_(1)) -(10^(-2))/(2) (T_(2)^(2)-T_(1)^(2))`
`DeltaG = 10 (1.5 - 1.0) xx 100 -2(300 -200) - (10^(-2))/(2)`
`(300^(2) - 200)^(2) = 50J [(P_(1))/(T_(1))=(P_(2))/(T_(2))rArrP_(2)=1xx(300)/(200)=1.5atm]`