Question

Two containers are filled with diatomic hydrogen gas and diatomic oxygen gas. The gases have the same temperature. Compare the average speed of hydrogen molecules to the average speed of oxygen molecules.

Answer 1

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Answer 2

Answer:

The ratio of the average speed of hydrogen molecules to the average speed of oxygen molecules, i.e., $V_{H_{2} } :{V_{O_{2} }$ is $4:1$ .

Explanation:

Given,

One container filled with diatomic hydrogen gas, i.e., $H_{2}$ gas

Another container filled with diatomic oxygen gas, i.e., $O_{2}$ gas

The temperature is the same.

The ratio of the average speed of the hydrogen molecules to the average speed of the oxygen molecules =?

As we know,

• The average speed ( $V_{rms}$ ) of the molecules can be calculated by the equation given below:
• $V_{rms} = \sqrt{\frac{2RT}{M} }$

Here,

• $V_{rms}$ = The average speed
• R = Raydberg's constant
• T = Temperature
• M = Molar mass

As given, the temperature is constant, therefore;

• $V_{rms} = \sqrt{\frac{1}{M} }$

Thus,

• $\frac{V_{H_{2} } }{V_{O_{2} } } = \sqrt{\frac{M_{O_{2} } }{M_{H_{2} } } }$             --------equation (1)

Now,

• The molar mass of $H_{2}$ = $2g/mol$
• The molar mass of $O_{2}$ = $32g/mol$

After putting the molar mass of the gases in equation (1), we get:

• $\frac{V_{H_{2} } }{V_{O_{2} } } = \sqrt{\frac{ 32 }{2}}$
• $\frac{V_{H_{2} } }{V_{O_{2} } } = \sqrt{16}$
• $\frac{V_{H_{2} } }{V_{O_{2} } } = 4$
• $V_{H_{2} } } = 4{V_{O_{2} } }$

Hence, the ratio of the average speed of hydrogen molecules to the average speed of oxygen molecules is $4:1$.