# Question #9b73a

##### 1 Answer

#### Answer:

The answer given in your book is correct.

#### Explanation:

As you know, this problem is nothing more than a straightforward application of the **van der Waals equation**

#color(blue)([p + a * (n^2/V^2)](V - nb) = nRT)" "# , where

**absolute temperature** of the gas

The interesting thing to notice here is that using the values given to you for

So, rearrange the above equation to solve for

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

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

This will give you

#p = (nRT - a(n^2/V^2)(V-nb))/(V-nb)#

#p = (nRT)/(V - nb) - a(n^2/V^2) color(red)(cancel(color(black)((V-nb))))/color(red)(cancel(color(black)((V-nb))))#

#p = (nRT)/(V - nb) - a(n^2/V^2)#

The units used for

#a = ["atm L"^2"mol"^(-2)]" "# and#" "b = "L mol"^(-1)#

Plug in your values to get

#p = (2color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 300color(red)(cancel(color(black)("K"))))/(5color(red)(cancel(color(black)("L"))) - 2color(red)(cancel(color(black)("moles"))) * 0.0371color(red)(cancel(color(black)("L")))color(red)(cancel(color(black)("mol"^(-1))))) - "0.17 atm" color(red)(cancel(color(black)("L"^2))) color(red)(cancel(color(black)("mol"^2))) * (2^2color(red)(cancel(color(black)("moles"^2))))/(5^2color(red)(cancel(color(black)("L"^2))))#

#p = "10.00 atm" - "0.0272 atm" = "9.97 atm"#

**However**, the *actual value* for the constant **not**

http://www2.ucdsb.on.ca/tiss/stretton/database/van_der_waals_constants.html

Redo the calculation using

#a = "4.17 atm L"^2"mol"^(-2)#

and you'll indeed get a pressure of

#p = color(green)("9.33 atm")#

As you can see, the problem here is that you were not given the correct value of