Kinetic Theory Of Temperature Problem

AI Thread Summary
The discussion revolves around calculating the temperature of a gas using the root mean square (rms) speed of its molecules. Given the rms speed of 600 m/s and the mass of the molecules at 4.6 x 10^-26 kg, the temperature is derived using the kinetic theory formula. The calculations lead to a temperature of 400 K, which converts to 127 °C. There are concerns regarding the accuracy of the calculations and the use of TeX formatting for equations. The importance of correctly applying the kinetic theory equations and understanding the relationship between rms speed, mass, and temperature is emphasized.
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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 Celcius?
<br /> 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&lt;v^2&gt;)\\<br /> t\=\frac{\frac{\2}{3}\(\frac{\1}{2}m&lt;v^2&gt;)}{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\\<br />
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\\
 
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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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