Question

Part C

15.0 moles of gas are in a 3.00 L tank at 23.1 ∘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.

Express your answer with the appropriate units.

Hints

pressure difference =

Answer 1

n = 15 mol

V = 3 L

T = 23.1C = 23.1+273 =296.1 K

find:

dP when --> methane is used

a)

IDEAL GAS LAW

PV = nRT

P = nRT/V

P = (15)(0.082)(296.1 )/(3)

P = 121.401 atm of methane (ideal gas law)

b)

for Wan deer Waals:

(P +a/v^2) * (v-b) = RT

where v = V/n

v = 3/15 = 0.2

solve for P.

(P +a/v^2) * (v-b) = RT

(P + 2.3/(0.2^2)) * (0.2-0.043) = 0.082*296.1

P = 0.082*296.1 / (0.2-0.043) - 2.3/(0.2^2)

P = 97.15 atm of methane, for vdw

c)

The pressure difference = 121.401 -97.15 = 24.251 atm difference deviation