At2 26.3C, F2 molecules have a certain value of rms speed. At what temperature would the rms speed of O2 molecules be the same?
To find the temperature at which the root-mean-square (rms) speed of O2 molecules would be the same as that of F2 molecules at 26.3°C, we need to apply the Kinetic Theory of Gases, which relates the rms speed of gas molecules to the temperature.
Let's use the equation for rms speed:
vrms = √(3kT/m)
Where:
- vrms is the root-mean-square speed of the molecules
- k is the Boltzmann constant (1.38 x 10^-23 J/K)
- T is the temperature in Kelvin
- m is the molar mass of the gas molecules
First, we need to convert the given temperature of 26.3°C to Kelvin:
T1 = 26.3°C + 273.15 = 299.45 K (rounded to two decimal places)
The molar mass of F2 is 2 * (19.00 g/mol) = 38.00 g/mol
The molar mass of O2 is 2 * (16.00 g/mol) = 32.00 g/mol
Now we can calculate the rms speed at T1 for F2:
vrms_F2 = √(3kT1/m_F2)
= √((3 * 1.38 x 10^-23 J/K * 299.45 K) / (38.00 g/mol * (1 kg/1000 g)))
≈ √(1.04 x 10^-21 J / (0.038 kg))
≈ √2.74 x 10^(-20) J/kg
≈ 1.65 x 10^3 m/s (rounded to two significant figures)
Now we need to find the temperature at which the rms speed of O2 molecules is the same as the calculated value for F2.
Let T2 be the temperature at which vrms_O2 = vrms_F2:
vrms_O2 = √(3kT2/m_O2)
√(3kT2/ (32.00 g/mol * (1 kg/1000 g))) = 1.65 x 10^3 m/s
Now we can solve for T2:
3kT2/(32.00 g/mol * (1 kg/1000 g)) = (1.65 x 10^3 m/s)^2
Simplifying:
T2 = [(1.65 x 10^3 m/s)^2 * (32.00 g/mol * (1 kg/1000 g))] / (3k)
= 861840 kg.m^2/s^2 / (3 * 1.38 x 10^-23 J/K)
≈ 6.05 x 10^5 K
So, at approximately 6.05 x 10^5 Kelvin, the rms speed of O2 molecules would be the same as that of F2 molecules at 26.3°C.