- #1
gsharples12
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The number density in a container of argon gas is 1.50×10e25 . The atoms are moving with an rms speed of 465 . What are (a) the pressure and (b) the temperature inside the container?
Vrms=(3KbT/m)^1/2
Na = 6.02X10^23 particles/mol
Ar=40u
p=F/A=(1/3)(N/V)mv^2
pV=NKbT
I started off with finding out how many moles there were in the container by taking 1.5e25 and dividing that by 6.02e23 and got 24.9 moles. I looked up Molar mass for Ar and got 40g per mol. The mass of the Ar is .9968 kg.
Then I am plugging what I found for m into the first equation to solve for T but the number for T doesn't look right. Am I on the right track?
Vrms=(3KbT/m)^1/2
Na = 6.02X10^23 particles/mol
Ar=40u
p=F/A=(1/3)(N/V)mv^2
pV=NKbT
I started off with finding out how many moles there were in the container by taking 1.5e25 and dividing that by 6.02e23 and got 24.9 moles. I looked up Molar mass for Ar and got 40g per mol. The mass of the Ar is .9968 kg.
Then I am plugging what I found for m into the first equation to solve for T but the number for T doesn't look right. Am I on the right track?