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

#PV=nRT#.

You are watching: Assuming it behaves as an ideal gas, calculate the density of helium, he, at stp.

Since we are in search of the density of #CO_2#, we can modify the legislation as follows:

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

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

Then reararray 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
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Truong-Son N.
May 18, 2016

REFERENCE DENSITY

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

Or, one deserve to calculate this from this website.

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

DENSITY ASSUMING IDEALITY

To get an principle of just how the density is like as soon as assuming ideality, we have the right to use the appropriate gas law to compare.

#mathbf(PV = nRT)#

where:

#P# is the pressure in #"bar"#. STP currently entails #"1 bar"# press.#V# is the volume in #"L"#.#n# is the #mathbf("mol")#s of gas .#R# is the global 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#, wbelow #M_m# is the molar mass of #"CO"_2# (#"44.009 g/mol"#, not #"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 from the true thickness, which is fairly good. Hence, #"CO"_2# is reasonably best.

DENSITY WITHOUT ASSUMING IDEALITY

Let"s calculate the thickness one more way.

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

From this website aget, I obtain #Z = 0.9934#.

Due to the fact that #Z , #"CO"_2# is easier to compress than a comparable ideal gas (for this reason its molar volume is much less than #22.711# at #"1 bar"# and also #"273.15 K"#).

Let"s see what its thickness 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))#

Hence...

#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 I did was use #Z# as a correctional factor in the ideal gas regulation. :)