Question

A monatomic gas is measured to have an average speed of 1477 m/s. If the
total amount of the gas is 2 mol (which equates to a mass of 0.008 kg), what
is the approximate temperature of the gas? (Recall that the equation for
kinetic energy due to translation in a gas is KEtranslational = 1 mv2 = 3 nRT,
2
and R = 8.31 J/(mol-K).)
2

Answer 1

Answer:

The temperature of the gas is 350.02 K.

Explanation:

The average speed is related to the temperature as follows:

$\bar v = \sqrt{\frac{3RT}{M}}$   (1)

Where:

$\bar v$: is the average speed = 1477 m/s

R: is the gas constant = 8.31 J/(K*mol)                                                  

T. is the temperature =?

M: is the molar mass

First, let's find the molar mass:

$M = \frac{m}{n}$

Where:

m: is the mass of the gas = 0.008 kg

n: is the number of moles = 2 mol

$M = \frac{m}{n} = \frac{0.008 kg}{2 mol} = 0.004 kg/mol$

Hence, by solving equation (1) fot T we have:

$T = \frac{\bar v^{2}*M}{3R} = \frac{(1477 m/s)^{2}*0.004 kg/mol}{3*8.31 J/(K*mol)} = 350.02 K$

Therefore, the temperature of the gas is 350.02 K.

I hope it helps you!