# Question #d132b

##### 1 Answer

#### Explanation:

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

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

where

#P# - the pressure of the gas

#V# - the volume it occupies

#n# - the number of moles of gas

#R# - theuniversal gas constant, usually useful as#0.0821 ("atm" * "L")/("mol" * "K")#

#T# - theabsolute temperatureof 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 **molar mass** of methane,

You will have

#n = m/M_M" " " "color(orange)((1))#

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

You will have

#rho = m/V " " " "color(orange)((2))#

Plug equation

#PV = m/M_M * RT#

This is equivalent to

#P * M_M = m/V * RT#

Now use equataion

#P * M_M = rho * RT#

This gets you

#color(purple)(bar(ul(|color(white)(a/a)color(black)(rho = P/(RT) * M_M)color(white)(a/a)|)))#

The molar mass of methane is equal to **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))))#

#color(green)(bar(ul(|color(white)(a/a)color(black)(rho = "1.7 g L"^(-1))color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.