Question

1.According to the ideal gas law, a 1.052 mol sample of methane gas in a 1.031 L container at 271.3 K should exert a pressure of 22.72 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For CH4 gas,

a = 2.253 L2atm/mol2 and
b = 4.278×10-2 L/mol.

Answer 1

They differ in a 5.86%

Step-by-step explanation:

Hi, the Van der Waals equation for gases is:

`(P+ (a)/(v^2))(v-b)=R*T`

For methane gas:

`a= 2.253 (L^2atm)/(mol^2)`

`b= 4.278*10^(-2) (L)/(mol)`

The conditions give{n are T=271.3 K, n=1.052mol and V=1.031L

The molar volume:

`v= (V)/(n)`

`v= (1.031L)/(1.052mol)`

`v= 0.98 (L)/(mol)`

Replacing in Van der Waals :

`(P+ (2.253 (L^2atm)/(mol^2))/( 0.98 (L)/(mol)^2))( 0.98 (L)/(mol)-4.278×10^-2 (L)/(mol))=0.082 (L*atm)/(mol*K)*271.3K`

`P=21.39 atm`

In percentaje:

`\frac{P_(VdW)}{P{ideal}}*100=(21.39 atm)/(22.72atm)*100`

`\frac{P_(VdW)}{P{ideal}}*100=94.14`

They differ in a 5.86%