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  • Find the temperature at which oxygen molecules would have the same root mean square (rms) speed as hydrogen molecules at <img alt="300 \mathrm{~K}" src="https://entrancecorner.oncodecogs.

Find the temperature at which oxygen molecules would have the same root mean square (rms) speed as hydrogen molecules at 300 \mathrm{~K}.

Option: 1

5400 \mathrm{~K}


Option: 2

4800 \mathrm{~K}


Option: 3

3200 \mathrm{~K}


Option: 4

1600 \mathrm{~K}


Answers (1)

best_answer

The root mean square speed (rms speed) of gas molecules can be calculated using the formula:

v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}

where, -v_{\text {rms }} is the rms speed of the gas molecule - k  is the Boltzmann constant \left(1.380649 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right. in S.I. unit )-T is the temperature in Kelvin.

m is the molar mass of the gas molecule in kg.

For Hydrogen \left(H_{2}\right) molecules, the molar mass m_{H_{2}} is approximately 2.016 \mathrm{~g} / \mathrm{mol} which is 32.00 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}.

Since we want the rms speed of oxygen molecules , \left(v_{\mathrm{rms}}, \mathrm{O}_{2}\right) 

to be equal to the rms speed of hydrogen molecules \left(v_{\mathrm{rms}}, \mathrm{H}_{2}\right)at 300 K. We can set up the following equation,

T_{O_{2}}=\frac{m_{O_{2}}}{m_{H_{2}}} \times T_{H_{2}}

Substitute the values:

 T_{O_{2}}=\frac{32.00 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}}{2.016 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}} \times 300 \mathrm{~K}

Calculating the value , T_{O_{2}}=4800 \mathrm{~K}.

The temperature at which oxygen molecules would have the same rms speed as hydrogen molecules at 300 K is approximately 48000 K.

So, the correct option is B

 

 

 

Posted by

vishal kumar

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