How many molecules of monotonic and diatomic gas are in a container?

  • Thread starter vjk2
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1. You have a container of 3300 cm^3. Half of it is diatomic oxygen and half is monotonic helium. Pressure is 17 atm and temperature is 25 degrees C. How many molecules of each substance are in it?



2. PV = 2/3 N(1/2 mv^2) -> PV = 2/3 N (K) -> N = 1.5 PV/K

K_monotonic = 3/2kT = 3/2 (1.38*10^-23)(298k)

k_diatonic = 5/2kT = 5/2 (1.38*10^-23)(298k)




3. N=1.5(17atm*(1Pa/9.869*10^-6 atm)*1650 cm^3 * 1m^3/10^6 cm^3))/((3/2) 1.38*10^-23)(298k))

3/2 factors cancel out. after all the math I get 6.91*10^23.

diatomic is basically 3/5 of that number, 4.15*10^23
 

Answers and Replies

  • #2
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masteringphysics is truly an awful piece of software. It turns out that the reason I was getting the wrong answer was because the gauge pressure undermeasures the pressure. It's supposed to be 18, not 17.
 

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