RMS unit conversion

Y9329124888

2014-03-11 8:46 pm

There is a question on the textbook, and I doubt the calculation steps stated.

Given sealed can of volume 350 cm^3 is full of air at pressure of 120 kPa, and temperature 20 C.
Assume that air behaves like an ideal gas, take R = 8.31 J per mol per K, average molar mass of air = 29.0 g per mol

Question in textbook:
Find the root-mean-square speed of the air particles in the can.

Answer in textbook:
SQRT(3RT / mNA) = SQRT( [3 x 8.31 x (20+273)] / (29xE-3) )

My Question is:
There is a step one in the question, which calculates the number of moles of air, and the answer is 0.0172 mol

Why mNA in the Answer in textbook uses figure of molar mass of gas of 29.0 g per mol? Rather than calculate mNA by Multiply 29.0 g per mol by 0.0172 mol to obtain the mass, and times NA constant of 6.02 x E23

To put it simply,
Why mNA = molar mass of gas?

----------------------------------------------
Reference:
N (No. of gas molecules) = n (no. of mol. of molecules) x NA (Avogadro Constant)
molar mass = mass of 1 mol of gas

回答 (1)

天同

2014-03-12 12:26 am

✔ 最佳答案

In your given formula: rms speed = SQRT(3RT / mNA)
here m is the mass of a molecule, and (NA) is Avogadro's Number.

Mass of a molecule is in unit of (kg), whereas Avogadro's number is in unit of "molecules/mole"..
Thus the product m(NA) is in unit of kg x molecules/mole = kg/mole, which is the mass of one mole of gas, and that is the Molar Mass of the gas.

2014-03-11 16:35:28 補充：
The calculation of 29 g (0.029 kg) x 0.0172 moles gives you the mass of air in that sealed can, not the mass of one mole of air. The rms speed would not be dependent on the mass of air in a particular can.

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