A vessel contains 1.60g of oxygen and 2.80g of nitrogen. The temperature is maintained at 300K and the voume of the vessel is `0.166m^(3)`. Find the pressure of the mixture.

Explanation

(P_(O_2)) = (n_(O_2))RT/ V`
` (P_(N_2)) = ((n_(N_2)RT)/V)`
` (n_(O_2)) = (m/(M_(O_2))) = 1.60/32 `
` = 0.05`
` (n_(N_2)) = (m/M_(N_2)) = 2.80/ 28 = 0.1`
` Now, (P_mix)= ((N_(O_2)) + (n(N_2))/V) RT`
` P_mix = ((0.05 xx 0.1) xx 8.3 xx 300 / 0.168)`
`= 2250 N/(m^2)`