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

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

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

$P = 98.041 - 40.563$

$P = 57.478atm$

Answer 2

The pressure of the chlorine gas according to the van der waal equation has been 13.76 atm.

Van der waal equation gives the idea that the gas reacts ideally.

The van der waal equation can be given by:

PV = $\rm \dfrac{RT}{V\;-\;b}\;-\;\dfrac{a}{V^2}$

where, P = pressure of the gas

V = volume of gas = 2 L

R = constant = 0.0821

T = temperature = 400 K

a = 6.49

b = 0.0652

Substituting the values

P $\times$ 2 = $\rm \dfrac{0.0821\;\times\;400}{2\;-\;0.065}\;-\;\dfrac{6.49}{(2)^2}$

2P = 27.53 atm

P = 13.76 atm,

The pressure of the chlorine gas according to the van der waal equation has been 13.76 atm.

For more information about the van der Waal's equation, refer to the link:

https://brainly.com/question/13201335