Question

At a pressure of 1 atm and at 115 K, what is the mass density (in grams per liter) of CH4 (boiling point = -161.5 C) according to:

a) the perfect gas equation of state (5 pts), and

b) the van der Waals equation of state (15 points).

Answer 1

from gas law, PV= nRT

n= mass/molar mass

PV= (mass/molar mass)*RT

P* Molar mass= (mass/Volume)* RT, R= gas constant =0.0821 L.atm/mole.K

for CH4, molar mass= 16 , mass/desnity= mass density

1* 16= mass density*0.0821*115

mass density= 16/(0.0821*115)= 1.69 g/L

2. for Vanderwaal equation of state

(P+an²/V²)*(V-nb)= nRT

take the basis as 1 mole,

the equation becomes (P+a/V²)*(V-b)= RT,

for CH4, a = 2.25 atmL2/mole2 and b= 0.0428 L/mol

hence (1+2.25/V²)*(V-0.0428)= 0.0821*115=9.4415

since it is non-linear equation, assume some value of V and match LHS and RHS with that value. If LHS does not match with RHS, assume some other value

The trail and error calculations gives a value of 52.5 L, the calculations are shown below

hence volume= 52.5 L

1 mole of CH4 means, mass= molar mass= 16gm

mass density= 16/52.5 g/L= 0.3047 g/L