Assuming the carbon dioxide behaves ideally, then we deserve to use the right gas law:

#PV=nRT#.

You are watching: Density of carbon dioxide at stp

Since us are searching for the thickness of #CO_2#, we have the right to modify the legislation as follows:

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

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

Then rearrange the expression to become:

#P=m/V(RT)/(MM)# wherein #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)#

Answer link

Truong-Son N.

may 18, 2016

**REFERENCE DENSITY**

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

Or, one have the right to calculate this native this website.

This also gives a **real fixed density** that #color(blue)("0.001951 g/mL")# in ~ #"1 bar"# and also #0^

"C"#.

**DENSITY suspect IDEALITY**

To get an idea of exactly how the thickness is like once assuming ideality, we deserve to use the **ideal gas law** come compare.

#mathbf(PV = nRT)#

where:

#P# is the**pressure**in #"bar"#. STP right now involves #"1 bar"# pressure.#V# is the

**volume**in #"L"#.#n# is the #mathbf("mol")#

**s that 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 exactly how #(nM_m)/V = rho#, where #M_m# is the molar fixed of #"CO"_2# (#"44.009 g/mol"#, no #"40 g/mol"#...), and also #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 native the true density, which is rather good. Thus, #"CO"_2# is reasonably ideal.

**DENSITY there is no ASSUMING IDEALITY**

Let"s calculation the density an additional way.

We can likewise use the **compressibility factor** #Z = (PV)/(nRT)#, which is one *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 gain #Z = 0.9934#.

Since #Z , #"CO"_2# is much easier to compress 보다 a comparable ideal gas (thus that is molar volume is less than #22.711# in ~ #"1 bar"# and #"273.15 K"#).

Let"s view what its density is this time.

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#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 in ~ that... It"s dead-on, and all ns did was use #Z# together a *correctional factor* in the appropriate gas law. :)