Question

What would be the pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container found using the van der Waals equation? For Cl₂, a = 6.49 L²･atm/mol² and b = 0.0652 L/mol.

Answer 1

Answer:

57.478atm

Explanation:

T = 400k

n = 5mol

v = 2.00

a = $6.49L^{2}$

b = 0.0652L/mol

R = 0.08206

Formula

P =  $\frac{RT}{(\frac{v}{n} )-b} - \frac{a}{(\frac{v}{n} )^{2} }$

$P = \frac{0.08206 * 400}{(\frac{2.00}{5.00} )-0.0652} - \frac{6.49}{(\frac{2.00}{5.00} )^{2} }$

$P = \frac{32.824}{0.3348} - \frac{6.49}{0.16}$  

$P = 98.041 - 40.563$

$P = 57.478atm$