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

1 Answer
Feb 23, 2018

#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#