At what temperature average speed of nitrogen will be same as the RMS speed of oxygen at 300 kelvin?
avg speed = v = sqrt(8RT/pi*M)
rrms = sqrt (3RT/M)
Solve v equation for T
Solve rms sped for T
Set v = vrms = to each other.
Substitute values and solve for T
Post your work if you gets stuck.
RMS = [158(Kelvin Temp/Mol Wt)^1/2]meters/sec
RMS O2 = 158(300/32)^1/2 m/sec
= 484 m/sec
Kelvin Temp for N2 RMS to equal O2 RMS @300K = M(RMS/158)^2
=28(484/158)^2 = 263 Kelvin
Check ...
RMS N2 = 158(263/28)^1/2 m/sec
=484 m/sec
To find the temperature at which the average speed of nitrogen (N2) is the same as the root mean square (RMS) speed of oxygen (O2) at 300 Kelvin, we can use the formula for average speed and RMS speed.
Average speed of a gas molecule can be calculated using the formula:
<𝑣> = (8𝑘𝑇/π𝑚)^(1/2)
where <𝑣> is the average speed, 𝑇 is the temperature in Kelvin, 𝑘 is the Boltzmann constant (1.38 x 10^-23 J/K), and 𝑚 is the molar mass of the gas.
RMS speed of a gas molecule can be calculated using the formula:
Vrms = (3𝑘𝑇/𝑚)^(1/2)
where Vrms is the RMS speed.
Now we can equate the average speed of nitrogen to the RMS speed of oxygen:
(8𝑘𝑇𝑛2/π𝑚𝑛)^(1/2) = (3𝑘𝑇𝑜2/𝑚𝑜)^(1/2)
Cancelling out constants and rearranging the equation, we get:
(𝑛𝑇𝑛/𝑚𝑛)^(1/2) = (𝑜𝑇𝑜/𝑚𝑜)^(1/2)
Now, let's substitute the values for nitrogen and oxygen:
Molar mass of nitrogen (𝑚𝑛) = 28 g/mol
Molar mass of oxygen (𝑚𝑜) = 32 g/mol
Plugging in the values, we get:
(𝑛𝑇𝑛/28)^(1/2) = (𝑜𝑇𝑜/32)^(1/2)
Next, let's square both sides of the equation to eliminate the square root:
𝑛𝑇𝑛/28 = 𝑜𝑇𝑜/32
Now, let's solve for 𝑛𝑇𝑛:
𝑛𝑇𝑛 = (𝑜𝑇𝑜/32) * 28
Finally, let's plug in the values:
Let's assume that the average speed of nitrogen is equal to the RMS speed of oxygen at a temperature 𝑇.
Therefore,
n * 𝑇 = (𝑇/32) * 28
Canceling out the temperature, we get:
n = 28/32
n ≈ 0.875
So, the average speed of nitrogen is equal to the RMS speed of oxygen at a temperature approximately equal to 0.875 times the temperature in Kelvin.