# Question d132b

Nov 1, 2016

${\text{1.7 g L}}^{- 1}$

#### Explanation:

Your starting point here will be the ideal gas law equation, which looks like this

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

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 useful as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Now, the idea here is that you need to use the fact that the number of moles of gas can be written in terms of the mass of a given mass, let's say $m$, and the molar mass of methane, ${M}_{M}$.

You will have

$n = \frac{m}{M} _ M \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Next, use the fact that the density of the gas is equal to the mass of the given sample, which we've said to be equal to $m$, and the volume it occupies, $V$.

You will have

$\rho = \frac{m}{V} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Plug equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ into the ideal gas law equation to find

$P V = \frac{m}{M} _ M \cdot R T$

This is equivalent to

$P \cdot {M}_{M} = \frac{m}{V} \cdot R T$

Now use equataion $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to write

$P \cdot {M}_{M} = \rho \cdot R T$

This gets you

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = \frac{P}{R T} \cdot {M}_{M}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The molar mass of methane is equal to ${\text{16.04 g mol}}^{- 1}$. Plug in your values to find -- do not forget to convert the temperature from degrees Celsius to Kelvin!

rho = (3.5 color(red)(cancel(color(black)("atm"))))/(0.0821 ( color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 127)color(red)(cancel(color(black)("K")))) * "16.04 g" color(red)(cancel(color(black)("mol"^(-1))))#

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = {\text{1.7 g L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.