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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.

1 Answer

4 votes

Answer:

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%

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