What is the density of carbon dioxide at STP?

2 Answers
May 17, 2016

Answer:

#d=1.79g*L^(-1)#

Explanation:

Assuming that carbon dioxide behaves ideally, then we can use the ideal gas law:

#PV=nRT#.

Since we are looking for the density of #CO_2#, we can modify the law as follows:

First we replace #n# by #n=m/(MM)# where, #m# is the mass and #MM=40g/(mol)# is the molar mass of #CO_2#.

#=>PV=nRT=>PV=(m)/(MM)RT#

Then rearrange the expression to become:

#P=m/V(RT)/(MM)# where #m/V=d# (#d# is the density).

#=>P=(dRT)/(MM)=>d=(PxxMM)/(RT)#

Therefore, #d=(1cancel(atm)xx40g/(cancel(mol)))/(0.08201(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))=1.79g*L^(-1)#

May 18, 2016

REFERENCE DENSITY

Wikipedia gives the density as #"0.001977 g/mL"# at #"1 atm"#, or if we convert it for #"1 bar"#, #color(blue)("0.001951 g/mL")#.

Or, one can calculate this from this website.

This also gives a real mass density of #color(blue)("0.001951 g/mL")# at #"1 bar"# and #0^@ "C"#.

DENSITY ASSUMING IDEALITY

To get an idea of how the density is like when assuming ideality, we can use the ideal gas law to compare.

#\mathbf(PV = nRT)#

where:

  • #P# is the pressure in #"bar"#. STP currently involves #"1 bar"# pressure.
  • #V# is the volume in #"L"#.
  • #n# is the #\mathbf("mol")#s of gas .
  • #R# is the universal gas constant, #"0.083145 L"cdot"bar/mol"cdot"K"#.
  • #T# is the temperature in #K"#.

#P/(RT) = n/V#

Notice how #(nM_m)/V = rho#, where #M_m# is the molar mass of #"CO"_2# (#"44.009 g/mol"#, not #"40 g/mol"#...), and #rho# is the mass density in #"g/L"#. Thus:

#color(blue)(rho) = (PM_m)/(RT)#

#= (("1 bar")("44.009 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K"))#

#=# #"1.94 g/L"#

#=# #color(blue)("0.001937 g/mL")#

That is about #0.72%# error from the true density, which is quite good. Thus, #"CO"_2# is fairly ideal.

DENSITY WITHOUT ASSUMING IDEALITY

Let's calculate the density another way.

We can also use the compressibility factor #Z = (PV)/(nRT)#, which is an empirical constant related to how easily #"CO"_2# responds to compression. If #Z = 1#, then #"CO"_2# is perfectly ideal.

From this website again, I get #Z = 0.9934#.

Since #Z < 1#, #"CO"_2# is easier to compress than a comparable ideal gas (thus its molar volume is less than #22.711# at #"1 bar"# and #"273.15 K"#).

Let's see what its density is this time.

#color(green)(Z) = P/(RT)V/n#

#Z/(M_m) = P/(RT)V/(nM_m)#

#= color(green)(P/(RTrho))#

Thus...

#color(blue)(rho) = (PM_m)/(RTZ)#

#= (("1 bar")("44.009 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K")(0.9934))#

#=# #"1.9507 g/L"#

#~~# #color(blue)("0.001951 g/mL")#

Oh look at that... it's dead-on, and all I did was use #Z# as a correctional factor in the ideal gas law. :)