# How many moles of gas are contained in 1.25 L at 303 K and 1.10 atm?

Jun 17, 2016

$\text{0.0553 moles}$

#### Explanation:

Your tool of choice here will be the ideal gas law equation, which 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

Now, take a look at the units used in the expression of the universal gas constant. The units you have for volume, temperature, and pressure must match those 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)("Kelvin, K")aaaaaaaaaacolor(green)(sqrt())
color(white)(aaaaaaacolor(black)("Atmospheres, atm")aaaaacolor(black)("Atmospheres, atm")aaaacolor(green)(sqrt())

Rearrange the ideal gas law equation to solve for $n$

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

Plug in your values to find

$n = \left(1.10 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 1.25color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 303color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0553 moles}} \textcolor{w h i t e}{\frac{a}{a}} |}}\right)$

The answer is rounded to three sig figs.