Question #9b73a
1 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
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
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