At what temperature does the rms speed of O2 molecules equal 475 m/s?
Recall that the formula for root mean square speed is given by
v,rms = (3RT/M)^(1/2)
where
R = universal gas constant = 8.314 J/mol-K
T = temperature in K
M = molar mass in kg/mol
We know that the molar mass of O2 is 32 g/mol or 0.032 kg/mol, and the v,rms is equal to 475. Substituting,
475 = (3 * 8.314 * T / 0.032)^(1/2)
475^2 = 779.4375 * T
T = 225625 / 779.4375
T = 289.47 K
In degree Celsius, T = 289.47 - 273 = 16.5 C
Hope this helps :3
The rms (root-mean-square) speed of gas molecules can be calculated using the following formula:
v_rms = √(3 * k * T / m)
Where:
v_rms is the root-mean-square speed
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 in kg
The molar mass of O2 is approximately 32 g/mol or 0.032 kg/mol.
Let's substitute the given value for v_rms (475 m/s) into the formula and solve for T:
475 = √(3 * (1.38 x 10^-23) * T / 0.032)
Squaring both sides of the equation:
475^2 = 3 * (1.38 x 10^-23) * T / 0.032
Simplifying:
225625 = 0.4254 * T
Dividing both sides by 0.4254:
T ≈ 530,404 K
Therefore, the temperature at which the rms speed of O2 molecules equals 475 m/s is approximately 530,404 Kelvin.
The root mean square (rms) speed of gas molecules is given by the equation:
v(rms) = √(3kT / m)
where:
- v(rms) is the root mean square speed of gas molecules
- k is the Boltzmann's constant (1.38 × 10^-23 J/K)
- T is the temperature in Kelvin
- m is the molar mass of the gas molecule
In this case, we are looking for the temperature (T) at which the rms speed of O2 (molar mass = 32 g/mol) molecules equals 475 m/s.
To find the temperature, we rearrange the equation:
T = (v(rms)^2 * m) / (3k)
Let's plug in the values:
T = (475^2 * 0.032) / (3 * 1.38 × 10^-23)
Calculating this expression, we find:
T ≈ 769100 K
Therefore, at a temperature of approximately 769100 Kelvin, the rms speed of O2 molecules would equal 475 m/s.