Calculate the pressure of 1.33 mol of argon gas in a 6.33 L container at 200°C using the van der waals equation

Answer 1

A little extra

If we are talking about an ideal gas, we would use PV=nRT

But at lower temperatures and high pressures, gases are less likely to behave ideally

At low temperature, gas particles that are usually far apart are now closer together and can feel the force of molecular attraction, and the pressure would be less than that predicted of an ideal gas.In Van Der Waals equation " a " will represent this adjustment.

At high pressures a gas will have a higher volume than expected. In Van Der Waals equation "b" will represent this adjustment.

Van Der Waals equation

(P+a (n/v)^2 ) X (V-nb)=nRT

For argon

a=1.35L^2 ATM/mol

b=.0322L/mol

Answer 2

Van der waals equation:

(p + a*n^2/V^2)(V - nb)-nRT

a=0.1355 Pa-m^6/mol^2, b=3.20*10^-5 m^3/mol, n=1.33 mol, V=6.33 L, T-200+ 273.15 K, R=8.3144 J/k-mol

1 L=0.001 m^3

place values into van der waals equation and solve for p.

(p + 0.1355*1.33^2/(0.00633)^2)(0.00633 - 1.33*3.20*10^-5)=1.33*8.3144*473.15

p=826180 Pa

826180 Pa*(1atm/1.01325*10^5 pa) is 8.15 atm