Question 9b58c

Apr 3, 2016

$4 \cdot {10}^{6} \text{Pa}$

Explanation:

All you have to do here is use the molar mass of argon to determine how many moles you get in that $\text{325-g}$ sample, then use the ideal gas law equation to find the pressure of the gas.

So, you know that argon has a molar mass of ${\text{39.9 g mol}}^{- 1}$, which tells you that one mole of argon has a mass of $\text{39.9 g}$.

Use this value to find the number of moles of argon present in your sample

325color(red)(cancel(color(black)("g"))) * "1 mole Ar"/(39.9color(red)(cancel(color(black)("g")))) = "8.145 moles Ar"#

The ideal gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant
$T$ - the absolute temperature of the gas

At this point, the most important thing to do is make sure that you know the units used in the expression of the universal gas constant.

In your case, you can use the value

$R = 8.31 \textcolor{w h i t e}{a} \left(\text{Pa" * "m"^3)/("mol" * "K}\right)$

http://www.cpp.edu/~lllee/gasconstant.pdf

This will allow you to plug the values given to you in the ideal gas law equation without doing any unit conversions.

As you can see, the pressure of the gas will be expressed in pascals, $\text{Pa}$. Rearrange the ideal gas law equation to solve for $P$

$P V = n R T \implies P = \frac{n R T}{V}$

Plug in your values to get

$P = \left(8.145 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles"))) * 8.31("Pa" * color(red)(cancel(color(black)("m"^3))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 298color(red)(cancel(color(black)("K"))))/(0.005color(red)(cancel(color(black)("m}}^{3}}}}\right)$

$P = \text{4,034,023 Pa}$

Since you only have one sig fig for the volume of the gas, you can only have one sig fig for the resulting pressure

$P = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4 \cdot {10}^{6} \text{Pa} \textcolor{w h i t e}{\frac{a}{a}} |}}}$