Question

The average speed of $H_2$ at $T_1$ K is equal to that of $O_2$ at $T_2$ K. The ratio $T_1:T_2$ is

• A. 1:16
• B. 16:1
• C. 1:4
• D. 1:1

Answer 1

The Correct Option is A

The average speed of a gas is given by the formula:

$v_{\text{avg}} \propto \sqrt{\dfrac{T}{M}}$ Where: - $v_{\text{avg}}$ is the average speed of the gas molecules, - $T$ is the temperature in Kelvin, - $M$ is the molar mass of the gas. 

We are told that the average speed of $H_2$ at $T_1$ is equal to the average speed of $O_2$ at $T_2$. 

So, according to the equation above: 
$\sqrt{\dfrac{T_1}{M_{H_2}}} = \sqrt{\dfrac{T_2}{M_{O_2}}}$ Squaring both sides: 
$\dfrac{T_1}{M_{H_2}} = \dfrac{T_2}{M_{O_2}}$ 

The molar mass of $H_2$ is 2 g/mol, and the molar mass of $O_2$ is 32 g/mol, so: 
$\dfrac{T_1}{2} = \dfrac{T_2}{32}$ 

Cross-multiplying: 
$T_1 \cdot 32 = T_2 \cdot 2$ 

Solving for the ratio of temperatures: 
$\dfrac{T_1}{T_2} = \dfrac{2}{32} = \dfrac{1}{16}$ So, the ratio $T_1 : T_2$ is: 

Answer: 1:16