# Question 9b73a

Jan 24, 2016

#### Explanation:

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

$\textcolor{b l u e}{\left[p + a \cdot \left({n}^{2} / {V}^{2}\right)\right] \left(V - n b\right) = n R T} \text{ }$, 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 $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - 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 $\text{9.97 atm}$.

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

$p \cdot \left(V - n b\right) + a \cdot \left({n}^{2} / {V}^{2}\right) \left(V - n b\right) = n R T$

$p \cdot \left(V - n b\right) = n R T - a \left({n}^{2} / {V}^{2}\right) \left(V - n b\right)$

This will give you

$p = \frac{n R T - a \left({n}^{2} / {V}^{2}\right) \left(V - n b\right)}{V - n b}$

$p = \frac{n R T}{V - n b} - a \left({n}^{2} / {V}^{2}\right) \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(V - n b\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(V - n b\right)}}}}$

$p = \frac{n R T}{V - n b} - a \left({n}^{2} / {V}^{2}\right)$

The units used for $a$ and $b$ should be

a = ["atm L"^2"mol"^(-2)]" "# and ${\text{ "b = "L mol}}^{- 1}$

Plug in your values to get

$p = \left(2 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 300color(red)(cancel(color(black)("K"))))/(5color(red)(cancel(color(black)("L"))) - 2color(red)(cancel(color(black)("moles"))) * 0.0371color(red)(cancel(color(black)("L")))color(red)(cancel(color(black)("mol"^(-1))))) - "0.17 atm" color(red)(cancel(color(black)("L"^2))) color(red)(cancel(color(black)("mol"^2))) * (2^2color(red)(cancel(color(black)("moles"^2))))/(5^2color(red)(cancel(color(black)("L}}^{2}}}}\right)$

$p = \text{10.00 atm" - "0.0272 atm" = "9.97 atm}$

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

$a = {\text{4.17 atm L"^2"mol}}^{- 2}$

and you'll indeed get a pressure of

$p = \textcolor{g r e e n}{\text{9.33 atm}}$

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.