Question

12.0 moles of gas are in a 4.00 L tank at 24.4 ∘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

Solution:

Given data ,

Number of moles of Gas (n) = 12moles

volme of tank (V) = 4 L

Temperature(T)= 24.4 degrees Celsius = 24.4+273=297.4K

Vander Waals constants for methane are

a = 2.300L2.atm/mol2

b= 0.0430L/mol

We know for ideal gas ,

PV=nRT

P= nRT/V

= (12*0.08206* 297.4)/4

=73.214atm

And we know Vander waals equation is,

(P+an²/V²)(V-nb) =nRT

P + (an²/V²) = nRT/(V-nb)

P =nRT/(V-nb) - (an²) /V²

P= (12*0.08206*297.4)/(4-(12*0.0430)) - (2.300*12²)/4²

^{= 63.357atm}

^{Therefore difference in pressure between ideal gas and methane is}

^{= 73.214 - 63.357 = 9.857atm}