# What pressure, in atmospheres, is exerted by 0.325 moles of hydrogen gas in a 4.08 L container at 35°C?

Jun 27, 2016

$\text{2.0 atm}$

#### Explanation:

This is a pretty straightforward example of how an ideal gas law problem looks like.

Your strategy here will be to use the ideal gas law to find the pressure of the gas, but not before making sure that the units given to you match those used by the universal gas constant.

So, the ideal gas law equation looks like this

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

Here you have

$P$ - 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

Take a look at the units given to you for the volume and temperature of the gas and compare them with the ones used in the expression of $R$.

$\textcolor{w h i t e}{a a a a a a a a a a a \textcolor{R e d}{\text{Need")aaaaaaaaaaaaaaacolor(blue)("Have}} a a a a a}$
$\textcolor{w h i t e}{a a a a a} \frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a}}$
color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaaaaaacolor(black)("Liters, L")aaaaaaaaaaacolor(green)(sqrt())
color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaaaaaacolor(black)("Celsius,"""^@"C")aaaaaaaaacolor(red)(xx)

Notice that the temperature of the gas must be expressed in Kelvin in order to work, so make sure that you convert it before plugging it into the ideal gas law equation

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 find

$P = \left(0.325 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (35 + 273.15)color(red)(cancel(color(black)("K"))))/(4.08color(red)(cancel(color(black)("L}}}}\right)$

$P = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{2.0 atm}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the temperature of the gas.