Assuming that carbon dioxide behaves ideally, climate we have the right to use the right gas law:

#PV=nRT#.

You are watching: Density of carbon dioxide gas at stp

Since we are trying to find the density of #CO_2#, we can 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 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)#

Answer attach

Truong-Son N.

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 have the right to calculate this from this website.

This additionally gives a **real fixed density** the #color(blue)("0.001951 g/mL")# at #"1 bar"# and also #0^

"C"#.

**DENSITY presume IDEALITY**

To get an idea of how the density is like when assuming ideality, we have the right to 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 the 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#, whereby #M_m# is the molar massive of #"CO"_2# (#"44.009 g/mol"#, no #"40 g/mol"#...), and also #rho# is the mass thickness 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 around #0.72%# error native the true density, i m sorry is quite good. Thus, #"CO"_2# is relatively ideal.

**DENSITY without ASSUMING IDEALITY**

Let"s calculate the density one more way.

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

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

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

Let"s check out 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 at that... It"s dead-on, and also all ns did was usage #Z# as a *correctional factor* in the right gas law. :)