The root mean square speed of fluorine gas at 425°C is 677 m/s.
FURTHER EXPLANATION
The root mean square speed of a gas is a measure of how fast gas particles move. It is affected by the molecular mass of the gas and the temperature.
To solve for the root mean square speed, [tex]v_{rms}[/tex] , the following equation is used:
[tex]v_{rms} \ = \sqrt{\frac{3RT}{M} }[/tex]
where:
R = gas constant = [tex]8.314 \ \frac{J}{mol - K} \ = 8.314 \ \frac{kg - m^2}{K-mol-s^2}[/tex]
T = temperature in Kelvin
M = molecular mass in [tex]\frac{kg}{mol}[/tex]
To find the root mean square speed of fluorine gas, sort the given information first:
T = 425°C
M of fluorine gas, [tex]F_2[/tex] = 38 g/mol
Ensure that the quantities are expressed with the correct units. In this case, the temperature must first be converted to K as follows:
[tex]T \ in \ K \ = ^oC \ + \ 273.15\\T \ in \ K \ = 425^oC \ + \ 273.15\\\boxed {T \ = 698.15 \ K}[/tex]
Next, the molecular mass of fluorine must be expressed in kg:
[tex]molecular \ mass \ in \ kg \ = given \ mass \ in \frac{g}{mol} \ (\frac{1 \ kg}{1000 \ g})\\molecular \ mass \ of \ F_2 \ = 38 \frac{g}{mol} \ (\frac{1 \ kg}{100 \ g})\\\boxed {molecular \ mass \ of \ F_2 \ = 0.038 \frac{kg}{mol}}[/tex]
Now that all values are expressed in the correct units, they may be plugged into the equation for [tex]v_{rms}[/tex]:
[tex]v_{rms} \ = \sqrt{\frac{3RT}{M} }\\v_{rms} \ = \sqrt{\frac{3(8.314 \frac{kg-m^2}{K-mol-s^2})(698.15 \ K)}{0.038 \frac{kg}{mol}}\\[/tex]
[tex]v_{rms} \ = \sqrt{458243.61 \frac{m^2}{s^2} }\\\boxed {v_{rms} \ = 676.93 \frac{m}{s}}\\[/tex]
Since the given in the problem has 3 significant figures only, then the final answer should be 677 m/s.
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Keywords: root mean square speed
