Calculating Pressure and Temperature in a Container of Argon Gas

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The discussion focuses on calculating the pressure and temperature of argon gas in a container with a number density of 1.50×1025 atoms/m3 and an rms speed of 465 m/s. The user successfully determined the number of moles of argon to be 24.9 by dividing the number density by Avogadro's number (6.02×1023 particles/mol) and calculated the mass of argon as 0.9968 kg. Using the formula for pressure, p = (1/3)(N/V)mv2, the user confirmed the correct pressure value, while seeking clarification on the temperature calculation using Vrms = (3KbT/m)1/2.

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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?
 
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So I figured out that 1 atom of Ar = 6.627E-26 and pluged it into p=(1/3)(N/V)mv^2
and got the correct answer for Pressure.
 

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