Question

Calculate the pressure exerted by Ar for a molar volume 0.45 L at 200 K using the van der Waals equation of state. The van der Waals parameters a and b for Ar are 1.355 bar dm mol-2 and 0.0320 dm3mol?, respectively. Please write your answer (unit: bar) with 2 decimals, as 12.23. Please do not add unit to your answer.

Answer 1

Step-by-step explanation:

It is known that the Van der Waals equation is a description of real gases, as in this equation there are all those interactions which we previously ignore in the ideal gas law.

In Vander Waals equation, the repulsion and collision, between molecules of gases are being considered. They are no longer ignored and they also are not considered a "point" particle.

According to the ideal gas law, PV = nRT

or,
`P((V)/(n))` = RT

and, let
`(V)/(n)` = v; which is called molar volume

Hence, P × v = RT

As, the van der Waals equation corrects pressure and volume as follows.

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

where, R = idel gas law; recommended to use the units of a and b; typically bar/atm and dm/L

T = absolute temperature, in K

v = molar volume, v =
`\frac{\text{Volume of gas}}{\text{moles of gas}}`

P = pressure of gas

Now, substitute the data in Vander Waal,s equation as follows.

For argon,
`(P+ (a)/(v^(2))) * (v - b)` = RT

`(P+ (1.355)/((0.45)^(2))) * (0.45- 0.0320)` = 0.08314
`bar dm^(3)/molK * (200)K`

(P+ 6.691) =
`0.08314 * (200)/((0.45- 0.0320))`

P = (39.7799 - 6.691) bar

P = 33.0889 bar

or, P = 33.09 bar (approx)

Thus, we can conclude that the pressure exerted by Ar in the given situation is 33.09 bar.