One mole of nitrogen is contained in a vessel of volume `V = 1.00 1`. Find :
(a) the temperature of the nitrogen at which the pressure can be calculated from an ideal gas law with an error `eta = 10 %` (as compared with the pressure calculated from the Van der Walls equation of state) ,
(b) the gas pressure at this temperature.

Explanation

(a) `p = [(RT)/(V_m - b) -(a)/(V_M^2)](1 + eta) = (RT)/(V_M)`
(The pressure is less for a Vander Wall gas than for an ideal gas)
or, `(a(1 + eta))/(V_M^2) = RT[(-1)/(V_M) + (1 + eta)/(V_M - b)] = RT (eta V_M + b)/(V_M(V_M - b))`
or, `T = (1(1 + eta)(V_M - b))/(R V_M(eta V_M + b))`, (here `V_M` is the molar volume)
=`(1.35 xx 1.1 xx (1-0.039))/(0.082 xx (0.139))~~ 125 K`
(b) The corresponding pressure is
`p =(RT)/(V_M - b) -(a)/(V_M^2) =(a(1 + eta))/(V_M(eta V_M + b))- (a)/(V_M^2)`
=`(a)/(V_M^2) ((V_M + eta V_M - eta V_M - b))/((eta V_M + b)) = (a)/(V_M^2) ((V_M - b))/((V_M + b))`
=`(1.35)/(1) = (0.961)/(0.139) ~~ 9.2 atm`.