Question

A sample of 0.440 mol of Ar gas has a volume of 6.55L and a pressure of 9.00 x 10^4 pa. What is the average velocity of the argon atoms? What is the temperature of the Ar?

Answer #1

**Solution:**

Volume **"V"** of gas = 6.55 L = **6.55 x
10 ^{-3} m^{3}**

Pressure **"P"** = **9 x10 ^{4}
Pa**

Number of moles **"n"** of argon = **0.44
mol**

Molar mass **'M"**of argon = 39.948 g/mol =
**39.948 x 10 ^{-3} kg/mol**

From kinetic theory of gases Pressur, volume and root mean square velocity "vrms"are related through the relation:

**PV = (1/3)nMv _{rms}^{2}** ; Where
M is the molar mass and n is the number of moles of a gas.

9 x10^{4} Pa x 6.55 x 10^{-3} m^{3} =
(1/3) x 0.44 mol x 39.948 x 10^{-3} kgmol^{-1} x
v_{rms}^{2}

**v _{rms} = 317.2 m/s**

We know that

v_{rms} = (3RT/M)^{1/2}

317.2 m/s = [3 x 8.314 JK-1mol-1 x T/(39.948 x 10^{-3}
kgmol^{-1})]^{1/2}

317.2 m/s = [ 624.36167 x T]^{1/2}

**T
= 161.15 K**

Hence, average velocity can be calculated from the following expression

v_{avg} = (8RT/M)^{1/2}

v_{avg} = [8 x 8.314 JK-1mol-1 x 161.15 K
**/**(39.948 x 10^{-3} kgmol^{-1} x
)]^{1/2}

**v _{avg} = 292.242 m/s**

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