Thermodynamics : molar specific heats for gases

[R2D2]

Homework Statement

An audience of 2750 fills a concert hall of volume 35000 m3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?

Homework Equations

Q= nCvΔT
Cv= (3/2) R

The Attempt at a Solution

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Since the v keeps constant, we use Cv to calculate the change of T of the air. We want to get the value of ΔT and we know n and Cv.

The total heat release by audience: 2750 x 70 x 2.0 x 60 x 60 = 1386000000 J

PV=nRT, so n = PV/RT = (1.013 x 10^5 x 35000) / (8.314 x (20 + 273.15)) = 1454713 mole

Cv = 1.5 x 8.314 =12.471

ΔT = Q/ (n x Cv) = 76.3 K... which doesn't make sense.

 

[Bystander]

What about your answer doesn't make sense?

 

[R2D2]

Oh sorry! I forgot to add that the answer is 43.1 K.

 

[Bystander]

(3/2)R is the constant volume heat capacity of a monatomic ideal gas. Give you any ideas?

 

[R2D2]

Yes I am kind of confused about which volume heat capacity I should use, since there are both monatomic and diatomic gas in air.
For diatomic gas, the heat capacity is 5/2 R, but I don't know if I should use that.

 

[SteamKing]

Yes I am kind of confused about which volume heat capacity I should use, since there are both monatomic and diatomic gas in air.
For diatomic gas, the heat capacity is 5/2 R, but I don't know if I should use that.

Which gases are monatomic? Nitrogen (79% of air)? Oxygen (21%)? Looks like everything else in the air is basically at trace levels only.

 

[R2D2]

Which gases are monatomic? Nitrogen (79% of air)? Oxygen (21%)? Looks like everything else in the air is basically at trace levels only.

So Cv should be 5/2 R instead of 3/2 R!
but I plug in Cv = 5/2 R I get 45.8 K as the answer, still not correct.. I don't know what is going wrong here with three degrees difference.

 

[R2D2]

anyone helps? : P