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)= 8T_(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))/(3xx 1.301-1)-(3)/((1.0301)^2)= 2.927-1.772= 1.1555, P= P_(r )P_(c )= 1.155xx 73.0atm = 84.35 atm`.