What is the density of a dioxygen gas at 840*"mm Hg"840⋅mm Hg pressure at a temperature of 291*K291⋅K?

1 Answer
anor277
May 5, 2017

rho<0.15*g*mL^-1ρ<0.15⋅g⋅mL−1

Explanation:

PV=nRTPV=nRT; i.e. P=(nRT)/V="mass"/Vxx(RT)/"molar mass"P=nRTV=massV×RTmolar mass.

And thus rho="mass"/V=P/(RT)xx"Molar mass"ρ=massV=PRT×Molar mass.

The signal problem with this question is that you do not measure pressure OVER 1*atm1⋅atm in mm*Hgmm⋅Hg. A mercury column is good for measuring pressures <=1*atm≤1⋅atm.

We know that 1*atm-=760*mm*Hg1⋅atm≡760⋅mm⋅Hg, i.e. an atmosphere will support a column of mercury that is 760*mm760⋅mm high......A mercury column is thus suitable to measure atmospheric pressure, or for pressures LESS than one atmosphere.

And thus, here, P=(840*mm*Hg)/(760*mm*Hg*atm^-1)=1.11*atm.P=840⋅mm⋅Hg760⋅mm⋅Hg⋅atm−1=1.11⋅atm.

And so, rho=(1.11*cancel(atm)xx32.0*g*cancel(mol^-1))/(0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx291*cancelK)

~=1.5*g*L^-1.........

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