Question #9b73a

1 Answer
Jan 24, 2016

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) = nRT)" "#, where

pp - the pressure of the gas
VV - the volume it occupies
nn - the number of moles of gas
RR - the universal gas constant, usually given as 0.0821("atm" * "L")/("mol" * "K")0.0821atmLmolK
TT - the absolute temperature of the gas
aa, bb - constants specific to every gas

The interesting thing to notice here is that using the values given to you for aa and bb will indeed result in a pressure of "9.97 atm"9.97 atm.

So, rearrange the above equation to solve for pp

p * (V - nb) + a * (n^2/V^2)(V-nb) = nRTp(Vnb)+a(n2V2)(Vnb)=nRT

p * (V - nb) = nRT - a(n^2/V^2)(V-nb)p(Vnb)=nRTa(n2V2)(Vnb)

This will give you

p = (nRT - a(n^2/V^2)(V-nb))/(V-nb)p=nRTa(n2V2)(Vnb)Vnb

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 and b should be

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 a for ammonia is 4.17, not 0.17.

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 a. The answer given in the book is correct, but part of the information provided is not.