`2 dm^(3)` of methane under `600 kPa` and `300K` and `4 dm^(3)` of oxygen under `900k Pa` and `300K` are forced into `3 dm^(3)` evacuated vessel, the temperature being maintained at `300K`. Calculate the change in entropy of gases assuming that they are ideal.

Explanation

Moles of `CH_(4) = (PV)/(RT) = (600 kPa xx 2dm^(3))/((8.314 kPa dm^(3)K^(-1))(300K))`
`= 0.481 mol`
Moles of `O_(2) = (PV)/(RT) = (900 xx4)/((8.314)xx300) = 1.443 mol`
`:. Delta_(mix) S = n_(A)"In"(V_(2))/(V_(1)) + n_(g)R "In" (V_(2))/(V_(1))`
SUbstituting values, we get
`Delta_(mix)S = (0.481) (8.314) In ((3)/(2)) +1.443 xx 8.314 xx In ((3)/(4))`
`= (1.6214 - 3.451) J K^(-1) =- 1.829 J K^(-1)`