Calculate the pressure exerted by one mole of `CO_(2)` gas at `50^(@)C`, confined to a volume of `0.125 dm^(3)`, using the law of corresponding states, given that the critical constants of the gas are `V_(m, c)= 0.0957 dm^(3), T_(c )= 304K" and "P_(c )= 73.0 atm`.

Explanation

According to principle of corresponding states `(P_(r) + 3//V_(r)^(2)) (3V_(r) - 1) = 8 T_(r)`
`P_(r) = (8T_(r))/(3V_(r) -1) = (3)/(V_(r)^(2)) = (8 (T//T_(c)))/(3 (V_(m)//V_(m , c)) - 1) - (3)/((V_(m) // V_(m , c))^(2)) = (8(323//304))/(3(0.125 //0.0957) - 1) - (3)/((0.125 // 0.0957)^(2)) = ((8) (1.0625))/(3 xx 1.301 - 1) - (3)/((1.301)^(2)) = 2.927 - 1.772 = 1.1555 , P = P_(r) P_(c) = 1.155 xx 73.0 "atm" = `84.35 atm