The molar mass of carbon dioxide is 0.045kg/mol. Calculate the density of the gas when the temperature is 273K and the pressure is 120000Pa..?

2 Answers
Dec 23, 2017

#rho_"carbon dioxide"=2.4*g*L^-1#

Explanation:

We assume ideality, and we know that #PV=nRT#

And so #P/(RT)=n/V=("mass"/"molar mass")/V#

And so #P/(RT)xx"molar mass"=underbrace("mass"/V)_("density", rho)#...

And so we plug in the numbers....

#rho_"carbon dioxide"=(120*kPa)/((273*Kxx8.31*L* kPa)/(K*mol))xx0.045*kg*mol^-1=2.38xx10^-3*kg*L^-1#

Dec 23, 2017

The density is #"2.3 kg/m"^3#.

Explanation:

The molar mass of #"CO"_2# is 0.044 kg/mol.

We can use the Ideal Gas Law to determine its density.

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

Since #n = m/M#, we can substitute this to get

#pV = (m/M)RT#

We can rearrange this to

#pM = m/VRT#

But #"density"= "mass"/"volume"# or #color(brown)(bar(ul(|color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "#

#pM = ρRT#

and

#color(brown)(bar(ul(|color(white)(a/a)ρ = (pM)/(RT)color(white)(a/a)|)))" "#

In your problem,

#p = "120 000 Pa"#
#M = "0.040 g/mol"#
#R = "8.314 Pa·m"^3·"K"^"-1""mol"^"-1"#
#T ="273K"#

#ρ = ("120 000" color(red)(cancel(color(black)("Pa"))) × 0.044 color(white)(l) "kg"·color(red)(cancel(color(black)("mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("Pa")))·"m"^3color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273color(red)(cancel(color(black)("K")))) = "2.3 kg/m"^3#