[iScience]

A problem wants me to calculate the mass of 1 mole of dry air which is a mixture of N₂(78% by volume), O₂ (21%), and Ar (1%)

i could do this if it was percentages by the percentage of the 1 mole, but i don't know where to start with volume percentages. don't tell me the entire thing but give me a starting point and hint to me where to go from there please.

thanks

[billnotgatez]

I got this from WIKI
http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure

The molar volume of any ideal gas may be calculated at various standard reference conditions as shown below:

    V/n = 8.3145 × 273.15 / 101.325 = 22.414 m³/kmol at 0 °C and 101.325 kPa
    V/n = 8.3145 × 273.15 / 100.000 = 22.711 m³/kmol at 0 °C and 100 kPa
    V/n = 8.3145 × 298.15 / 101.325 = 24.466 m³/kmol at 25 °C and 101.325 kPa
    V/n = 8.3145 × 298.15 / 100.000 = 24.790 m³/kmol at 25 °C and 100 kPa

[sjb]

A problem wants me to calculate the mass of 1 mole of dry air which is a mixture of N₂(78% by volume), O₂ (21%), and Ar (1%)

i could do this if it was percentages by the percentage of the 1 mole, but i don't know where to start with volume percentages. don't tell me the entire thing but give me a starting point and hint to me where to go from there please.

thanks

What would be the mass if it were 100% nitrogen. Perhaps also look at things like the ideal gas law?

[iScience]

so can you tell me if this is right?

0.78V_N2+0.21V_O2+0.01V_Ar=V_total

PV_total=nRT

V_total=(nRT/P)

0.78V(n_N2RT/P)+0.21(n_O2RT/P)+0.01(n_ArRT/P)=V_total

=(RT/P)(0.78n_N2+0.21n_O2+0.01n_Ar)=V_total

but i know since the total n must equal one, as given in the problem,

this becomes the equivalent problem as if i were doing percentages with moles right?

so..

(0.78n_N2+0.21n_O2+0.01n_Ar)=1

am i on the right track so far?

[Borek]

Yes.

[iScience]

thank you very much everyone

[billnotgatez]

@iScience
If you give your final steps I would be grateful