Kinetic Theory Of Temperature Problem

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SUMMARY

The discussion centers on calculating the temperature of a gas using the kinetic theory of temperature. Given the root mean square (rms) speed of the gas molecules at 600 m/s and the mass of the molecules at 4.6 x 10^-26 kg, the temperature was determined to be 400 K, which converts to 127 °C. The calculations utilized the equation T = (2/3)(1/2 m ) / R, confirming the relationship between molecular speed and temperature. The user expressed uncertainty about the correctness of their calculations and the formatting of their LaTeX code.

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Iscariot
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The rms speed of the molecules in a gas is 600ms^-1 and the mass of the molecules is 4.6*10^-26kg. What is the temperature of the gas in degrees celsius?
[tex] v_{r.m.s.}\=\600\ms^-1 \\<br /> m\=\4.6x10^-26kg \\<br /> n\=\6.022x10^23 \\<br /> r\=8.31\j\k\mol^-1 \\<br /> \\<br /> \\<br /> rt\=\frac{\2}{3}\(\frac{\1}{2}m<v^2>)\\<br /> t\=\frac{\frac{\2}{3}\(\frac{\1}{2}m<v^2>)}{r}\\<br /> t\=\frac{\frac{\2}{3}\(\frac{\1}{2}\x\4.6x10^-26\x\600^2)}{8.31}\\<br /> t\=\frac{3324.14}{8.31}\\<br /> t\=\400K\\<br /> t(c)\=\400-273\=\127c\\[/tex]
Does this seem correct? I have no way of checking the answer and I feel really unsure, especially using the Vr.m.s.
Thanks
EDIT: I can't understand why my TEX doesn't work :(

My tex code:

V_{r.m.s.}\=\600\ms^-1 \\
m\=\4.6x10^-26kg \\
N\=\6.022x10^23 \\
R\=8.31\J\K\mol^-1 \\
\\
\\
RT\=\frac{\2}{3}\(\frac{\1}{2}m<v^2>)\\
T\=\frac{\frac{\2}{3}\(\frac{\1}{2}m<v^2>)\\}{R}\\
T\=\frac{\frac{\2}{3}\(\frac{\1}{2}\x\4.6x10^-26\x\600)\\}{8.31}\\
T\=\frac{3324.14}{8.31}\\
T\=\400K\\
T(C)\=\400-273\=\127C\\
 
Last edited:
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You need to write the tex tag using lowercase letters.
 
1/2m(rms) = 3/2RT or possibly 1/2m(rms) = 3/2kT depends on the amount of gas molecules for one mole it would be the first one i think.
 

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