What is the density of carbon dioxide at STP?

2 Answers
May 17, 2016

d=1.79g*L^(-1)d=1.79gL1

Explanation:

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

PV=nRTPV=nRT.

Since we are looking for the density of CO_2CO2, we can modify the law as follows:

First we replace nn by n=m/(MM)n=mMM where, mm is the mass and MM=40g/(mol)MM=40gmol is the molar mass of CO_2CO2.

=>PV=nRT=>PV=(m)/(MM)RTPV=nRTPV=mMMRT

Then rearrange the expression to become:

P=m/V(RT)/(MM)P=mVRTMM where m/V=dmV=d (dd is the density).

=>P=(dRT)/(MM)=>d=(PxxMM)/(RT)P=dRTMMd=P×MMRT

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. :)