Skip to main content
Ch 18: Thermal Properties of Matter

Chapter 18, Problem 18

Martian Climate. The atmosphere of Mars is mostly CO2 (molar mass 44.0 g/mol) under a pressure of 650 Pa, which we shall assume remains constant. In many places the temperature varies from 0.0°C in summer to -100°C in winter. Over the course of a Martian year, what are the ranges of (a) the rms speeds of the CO2 molecules and

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
1069
views
1
rank
Was this helpful?

Video transcript

Welcome back everybody. We are looking at oxygen inside a container with a movable piston. We know that the molar mass of oxygen is 32 g a mole. And we are told that as this piston moves the temperature is going to fluctuate between negative 70 degrees Celsius and 40 degrees Celsius. And we are tasked with finding what is the range of the root mean square speeds of the oxygen molecules. So a K. We are going to need to find an upper bound for this speed and a lower bound for this speed. Well, the formula for root mean square speeds is given as follows, it is the square root of three times the ideal gas constant times are absolute temperature. A kR temperature in kelvin divided by our molar mass. So what we're gonna do is we're going to use each of these bounds of the temperature, plugging it into the equation twice to get our bounds for our speeds. Before we can start plugging in numbers, we need to make sure we're working in the correct units. So for our moller mass, we are given that it is 32 g per one mole. But we need to be working in kilograms per mole. So I'm gonna convert real quick. There are 1000 g in one kg. So I multiplied by such. These units will cancel out and we are left with that. Our molar mass is 32 times 10 to the negative third kilograms per mole. Now, as for our range of temperatures, we need to be dealing in absolute temperatures, which means we need to be dealing in kelvin. So what I'm gonna do is I'm gonna add 273 to each side of this. In equality to get our kelvin temperatures. This makes the lower end of temperature 203 kelvin and the upper end of temperature 3 13 kelvin. And I do apologize. I might just misspoke we are adding to 73 to both sides of our inequality to get our temperatures in kelvin. Wonderful. Now that we have all of that, we are ready to go ahead and plug into our formula right here. So first let's start with the lower root mean square speed. This is going to be equal to the square root of three times our ideal gas constant of 8.314 times our lower temperature bound of Calvin, all divided by our molar mass of oxygen of times 10 to the negative third. When you plug all this in your calculator, this gives us a lower root mean square speed of 398 m per second. Great. Now we're going to do the same thing for the upper bound here we have that. The upper bound is equal to the square root of three times our ideal gas constant of 8.314 times this time our upper bound of temperature 3, 13 Kelvin all divided by our molar mass of 32 times and to the negative third. Okay, this into our calculator, we get an upper bound of 494 m per second. Let me go ahead then and plug these values into this inequality here. And we will get that our root mean square speed is greater than or equal to 398 m per second, and it is less than or equal to 494 m per second, of course, corresponding to our final answer. Choice of the Thank you all so much for watching. Hope this video helped. We will see you all in the next one.