On earth, STP is based on the average atmospheric pressure at the surface and on a phase change of water that occurs at an easily produced temperature, being only slightly cooler than the average air temperature. The atmosphere of Venus is almost entirely carbon dioxide (CO₂), the pressure at the surface is a staggering 93 atm, and the average temperature is 470℃. Venusian scientists, if they existed, would certainly use the surface pressure as part of their definition of STP. To complete the definition, they would seek a phase change that occurs near the average temperature. Conveniently, the melting point of the element tellurium is 450℃. What are (a) the rms speed and (b) the mean free path of carbon dioxide molecules at Venusian STP based on this phase change in tellurium? The radius of a CO₂ molecule is 1.5 x 10⁻¹⁰ m.

Question

Accepted Answer

Hi everyone in this practice problem, we are being asked to calculate the mean free path and also the root mean square velocity. We will have an atmosphere of a terrestrial exoplanet composed primarily of nitrogen. The temperature and pressure at the surface of the exoplanet are thought to be 430 Calpine and 15 bars respectively. The planetary scientists studying this exoplanet define the standard temperature and pressure S D P as 430 kin and 15 bars. And we are being asked to calculate first the mean free path and second the root mean square velocity of nitrogen molecules on the planet's surface. We are being asked to assume that the ideal gas law is still applicable to this uh environment and the two nitrogen atoms are separated by a distance of 1.3 times 10 to the power of negative 10 m. The options given for the mean free pad and the root mean square velocities are a five nanometer and 370 m per second. B five nanometer and 620 m per second. C 13 nanometer and 370 m per second. And lastly D 13 nanometer and 620 m per second. So let's start with the first part which is the mean free path. So the mean free path is the average distance traveled by a moving molecule between collisions and the mean free path. For the first part of this problem can be calculated by this lambda which will have the formula of one divided by four, multiplied by squared of two multiplied by pi multiplied by open parentheses and divided by V close parenthesis multiplied by R squared N is the total number of nitrogen molecules occupying volume of P. So P is the volume occupied and R squared is the radius of the nitrogen molecules which are going to be assumed to be spherical. So in this case, it is also given that um the ideal gas law is still going to be applicable. So we are going to apply the ideal gas law on the exoplanet by uh using P P to be equals to N multiplied by K B multiplied by T. We are interested to find the N divided by V which is included in our lambda equation. So that's rearranged this equation that we have from the ideal gas law to obtain an equation for N divided by V. So then N divided by V will come out to B B divided by K B multiplied by T just like some. So we want to combine this two equation or substitute the N divided by V into the Lambda equation. So that we get Lambda to then be equal to one divided by four, multiplied by squared of two multiplied by pi multiplied by B divided by K BT multiplied by R squared just like sub. So substituting this with all of the values that we know from the problem statement, we can then actually calculate our mean free path. So let's start doing that. So lambda will then be equal to one divided by, we have four, multiplied by square root of two, multiplied by Pi multiplied by B divided by K B and T. So P is the pressure and it is given to be 15 bars. So we have to convert 15 bars into pascal because it is in uh pascal is the SI unit. So we have to multiply that with 10 to the power of five pascals for every one bar just like. So and then we want to divide the pressure with K B and P. So K B is just the Boltzmann constant which is 1. times 10 to the power of negative 23 jo divided by Kelvin. And then the temperature is given to be Kelvin. So since it is already in si, which is in Kelvin, we don't need to convert anything. And then the R squared is given to be 1.3 times 10 to the power of negative 10 m, which is the distance of separation between the two nitrogen atoms. So 1.30 times 10 to the power of negative 10 m squared just like so awesome. So then doing all of this calculation, we will then get the mean free pad to be nanometer. So that will be the answer to the first question. So let's move on to the first part of this problem statement which is asking us to get the R MS velocity. So the R MS velocity is directly proportional to the square root of temperature and inversely proportional to the square root of the molar mass or V R MS will be equal to the square root of three K B multiplied by T divided by M. So the mass of A N two molecule or nitrogen molecule is going to be equals to 28 U. So 28 U is in at atomic mass. So we have to convert that into kilogram in si. So we want to multiply that by 1.661 times 10 to the power of negative 27 kg for every one atomic mass unit or one U. So then calculating this uh nitrogen mass or nitrogen molecule mass, we can then get M to be equals to 5.65 times 10 to the power of negative 26 kg. So next, we want to substitute all of our values into our V R MS formula. So that V R MS will then equal to the square root of three K B which is three times 1.38 times 10 to the power of negative 23 Jules divided by Kelvin multiplied by the temperature which is going to be 430. Kelvin divide that with the mass atomic mass, not the mass, which is going to be 4.65 times 10 to the power of negative 26 kg just like So, so doing all this calculation, then we will get our V R MS value to then be equal to 618.7 m per second. So there we have it, the mean free path is going to be 30 nanometer and the R MS speed or R MS velocity. If we route it up is going to be 620 m per second. So there we go. Um all of these answers will actually correspond to option D in our answer choices, which will be the answer to this practice problem. So that will be it for this video with the mean free pad being nanometer and the R MS speed rounding to two significant figures being 620 m per second. That will be it for this video. And if you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that'll be it for this one. Thank you?

Verified Solution

Key Concepts

Root Mean Square Speed (rms speed)

Mean Free Path

Phase Change