Question

Question #9b73a

Answer 1

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

`p` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas
`a`, `b` - constants specific to every gas

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

So, rearrange the above equation to solve for `p`

`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` 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.