# What is the density of acetylene gas measured at pressure of 0.75 atm and temperature of 250°C?

Jan 5, 2016

$\rho = \text{0.45 g/L}$

#### Explanation:

Your strategy here will be to use the molar mass of acetylene, ${\text{C"_2"H}}_{2}$, and the ideal gas law equation to find an expression for the density of the gas under those conditions for pressure and temperature.

So, the ideal gas law equation looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, 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 temperature of the gas, always expressed in Kelvin

Now, since you don't know how many moles of acetylene you have, you will have to work around it using the definition of molar mass.

A substance's molar mass tells you what the mass of one mole of that substance is. You can thus write molar mass, ${M}_{M}$, as being equal to

$\textcolor{b l u e}{{M}_{M} = \frac{m}{n}} \text{ }$, where

$m$ - the mass of a sample of the substance
$n$ - the number of moles that sample contains

This means that you can express the number of moles, $n$, as being

${M}_{M} = \frac{m}{n} \implies n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

PV = m/M_M * RT" " " "color(red)(("*"))

Now, density, $\rho$, is defined as mass per unit of volume.

$\textcolor{b l u e}{\rho = \frac{m}{V}} \text{ }$, where

$m$ - the mass of a sample of the substance
$V$ - the volume it occupies

Rearrange equation $\textcolor{red}{\left(\text{*}\right)}$ to get the expression for density

$P \cdot {M}_{M} = {\overbrace{\frac{m}{V}}}^{\textcolor{b l u e}{= \rho}} \cdot R T$

This means that you have

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

Acetylene has a molar mass of $\text{26.037 g/mol}$. Plug in your values and solve for $\rho$ - do not forget to convert the temperature from degrees Celsius to Kelvin

rho = (0.75 color(red)(cancel(color(black)("atm"))) * 26.037"g"/color(red)(cancel(color(black)("mol"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 250)color(red)(cancel(color(black)("K")))) = "0.4547 g/L"

Rounded to two sig figs, the number of sig figs you have for the temperature and pressure of the gas, the answer will be

$\rho = \textcolor{g r e e n}{\text{0.45 g/L}}$