Question

Calculate the density of CH4 at 0°C and 1 atmospheric pressure?

Answer 1

`0.72"g/dm"^3`

Explanation:

We have The Ideal Gas Law:

`PV=nRT`

Since `V=m/rho`, where `m` is the mass of the gas and `rho` its density, and `n=m/M`, where `M` is the molar mass of the gas, we can write:

`(Pm)/rho=(mRT)/M`

We now have to solve for `rho`. Dividing both sides by `Pm` gives:

`1/rho=(mRT)/(MPm)`

Which reduces to:

`1/rho=(RT)/(MP)`

Reciprocate both sides:

`rho=(MP)/(RT)`.

Here,

`M=16.04"g/mol"`

`P=1"atm"`

`T=273.15"K"`

`R=0.0821"L atm K"^-1"mol"^-1`.

Inputting:

`rho=(16.04*1)/(0.0821*273.15)`

`rho=0.72`

Now we have our number. What about the unit?

We know that density's units are a mass unit divided by a volume unit.

The mass unit here is grams, as the molar mass was expressed in `"g/mol"`. The volume unit is liters, as the value of `R` contained that value. As `1"L"=1"dm"^3`, we can write the density as:

`0.72"g/dm"^3`