The rms speed of the atoms in a 2.10g sample of helium gas is 760m/s . What is t

ID: 1315677 • Letter: T

Question

The rms speed of the atoms in a 2.10g sample of helium gas is 760m/s . What is the thermal energy of the gas?

I figured I should solve for Temperature first, so I used the following formulas:

Vrms = sqrt(3*KB*T/m) => Vrms2*m/3*KB=T

Plugging in, I got (760 m/s)2*(0.0021 kg)/(1.38*10-23 J/K * 3) = 2.93*1025 K.

I thought that looked quite large to say the least, but it's the number the equation from the book gave me so I chugged along.

Eth = 1.5*nRT

2.1g He * 1mol He/4g He = 0.525 mol He = n

=> E th = 1.5*(0.525mol)(8.31 J/mol K)(2.93*105 K) = 1.92*1026 J

However, the online homework told me that answer is wrong.

What am I doing wrong? It seems like it should be a straightforward problem but I'm just not getting it.

Thanks in advance

You are getting the value of 'm' wron in the calculation of Temperature.

We know that atomic mass of Helium gas = 4 gm/mole = 4 * 10-3 kg/mol

The no. of atoms in one mole of any element = 6.022 x 1023 (Avagadro's number)

Therefore the atomic mass of He for one atom = 4 * 10-3/ (6.022 x 1023) = 6.6 * 10-27 kg

Pluggin in the above values, you get (760 m/s)2*(6.6 * 10-27 kg)/(1.38*10-23 J/K * 3) = 92.67K.

From here on it's straight forward.

Eth = 1.5*n*R*T

Eth = 1.5 * 0.525* 8.31 * 92.67 = 606.45 J

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