Question

13.0 moles of gas are in a 7.00L tank at 25.0?C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2?atm/mol2 and b=0.0430 L/mol.

Answer 1

A/c to ideal gas law

PV = nRT

ideal gas molecule behave likes a point particle and collison between them is prefectly elastic.

P= nRT/V

n=13, V=7L T= 273+25 =298 K

for ideal gas,

P= 13x0.0820x298 /7 =45.38 atm

Van der Wall's take into account moleculer size & moleculer interaction and gave aequation called Van der Wall equation which is given as

[P+a(n/V)^2](V/n-b)= RT

a=2.3 L2/atm/mol2 b= 0.043L/mol

[P+ 2.3(13/7)^2](7/13 -0.043) = 0.082x 298

[P+7.931](0.53846-0.043)=24.436

[P+7.931]= 24.436 / 0.49546=49.32

P= 49.32 - 7.931= 41.389 atm= Pressure of methane

Difference between pressure of ideal gas & methane gas

45.38 - 41.389 = 3.991 atm