Question

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N
2
​
. Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy. If the root-mean-square speed of the gas molecules is 182 m/s, what is the pressure of the gas?

Answer 1

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

Where:

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

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

Where:

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$    

Where:

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$: is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$        

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

Answer 2

The pressure of the gas is 0.015 atm if the gas obeys the ideal gas law.

We know that;

$vrms = \sqrt{\frac{3RT}{M}$

For N2;

vrms =  182 m/s

R = 8.314 J/K.mol

T = ?

M = 28 g/mol or 0.028 Kg/mol

So;

T = $\frac{vrms^{2} M}{3R}$

T = $\frac{(182)^{2} * 0.028}{3 *8.314}$

T = 37.2K

From PV = nRT

n = 0.226 g/28 g/mol = 0.0081 moles

P = nRT/V

P =  0.0081 moles × 0.082 atmLK-1mol-1  ×  37.2K/1.64 L

P = 0.015 atm

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