At what temperature (°C) will xenon atoms have the same average speed that Br2 molecules have at 20° C?

Video transcript

Welcome back everyone. In this example, we need to calculate the temperature wherein our sulfur dioxide molecules will have the same mean speed as our nitrogen dioxide molecules at 63 degrees Celsius, report the answer in degrees Celsius. So what we're going to be using for this prompt is our formula for the main square speed velocity, which we should recall is calculated by taking the square root of three, multiplied by r gas constant. R multiplied by r temperature in kelvin divided by the molar mass of our gas. So we should recall that R r gas constant is a value of 8. joules divided by moles times kelvin. Our temperature is again reported in the units of kelvin. Our molar masses reported in units of kilograms per mole and our root mean square speed or velocity is reported in units of meters per second. We also want to keep in mind that one jewel is equivalent to one kg times meters squared, divided by seconds squared. So what we're going to first do is calculate the root mean square speed of our nitrogen dioxide gas molecule. So we're gonna plug this into our formula so that we have the proper molar mass and temperature units. So we would recall that from our periodic table are molar mass for nitrogen dioxide for every atom in this molecule Is going to include our molar mass of nitrogen, which we see is located in group five a. And corresponds to a mass of 14.01 g per mole. This is then added to our Mueller massive oxygen, which we find in group six a on the periodic table corresponding to a mass of 16.00 g per mole. But we have that subscript of two next to our oxygen. So we're going to multiply this by two. So by two oxygen atoms to be specifically. And this is going to yield a molar mass of 64 Or sorry, 46 .01 g per mole. Now, as we stated from the prompt, we need molar mass for our mean square speed formula to be in kilograms per mole. So we're going to convert from grams to kilograms by recognizing that one g is equivalent to 10 to the negative third power kilograms. So now getting rid of grams, we can now be left with kilograms per mole as our final units of molar mass. And this is going to yield 0.4601 kg per mole as our molar mass of nitrogen dioxide. And now we want to get that temperature in our units of kelvin. It's given in the prompt as 63 degrees Celsius for the nitrogen dioxide gas molecule. And we're gonna recall that we add to 73.15 to get our kelvin temperature of 336.15 kelvin. So now finding that root mean square speed of our nitrogen dioxide gas, we're going to set this equal to the square root of three, multiplied by r gas constant R 8.314 joules divided by moles times kelvin multiplied by our temperature of 336.15 kelvin Divided by our molar mass of 0. kg per mole. And so focusing on simplifying everything, We're going to say that our root mean square speed of nitrogen dioxide is equal to our square root of 8384.25. And we're going to calculate our units of jewels or rather interpret our units of jewels as kilograms times meters squared, divided by seconds squared, divided by moles times kelvin multiplied. Or rather the kelvin unit. We can actually cancel out because it's canceled out with our kelvin temperature there. And now we're just going to divide by our denominator of 0. kg per mole. So simplifying this, our root mean square speed of nitrogen dioxide is now equal to 426.88 02. And for our units we can get rid of our meters squared to be left with one unit of meters and our second square to be left with one unit of seconds, we can cancel out kilograms as well as moles and our final units are going to be meters per second. So now that we have this mean square speed of nitrogen dioxide gas, we can use it to calculate our Temperature, where are sulfur dioxide gas has the same mean square speed as nitrogen dioxide. So taking our answer, we're going to say that 426. m/s is set equal to the square root of and sorry, let's fix this square root. So the square root of three times r gas constant R 8.314 joules divided by moles times kelvin multiplied by our temperature in as a variable T in kelvin and now divided by our molar mass of sulfur dioxide. So we can make note of our molar mass of sulfur dioxide by calculating that sulfur found in Group six A of our periodic table has a molar mass of 32.0 66 g per mole. And adding to this, we have our molar mass of oxygen also found in Group six A corresponding to 16.0 g per mole. Multiplied by R two oxygen atoms. And just so everything is visible. Let's scoot this over. So multiplied by our two oxygen atoms. Here, we're going to have a molar mass equal to 64.066 g per mole. But as we stated, we need this to be in kilograms per mole. So we're going to multiply again by and sorry, we need to multiply by kilograms over grams. Where one g is equal to 10 to the negative third power kilograms and canceling out grams, we're left with kilograms per mole as our molar mass units. So we're going to have 0.6. And let's actually write this below. So we have enough room. We'll have 0. kg per mole as our molar mass of sulfur dioxide gas, which we can now plug into the denominator. So 0.064066 kg per mole. So beginning by taking both sides to a power of two to get rid of that square U squared power term Or a sorry square root term. We're now going to simplify in our next line so that we have 426 or sorry, rather once we square, it we're going to have 182,226. meter squared divided by seconds squared set equal to our right hand side. Where we now have just 24.9 42. When we take the product of our numerator will have units which will now interpret where jewels are kilograms, times meters squared divided by seconds squared divided by moles, times kelvin multiplied by our temperature. T then divided by our denominator. 0.64066 kg per mole. So now focusing on canceling outer units, we can cancel out our units of moles as well as kilograms. And in our next line we're just going to divide so that we have 182, 226.715 m squared divided by seconds squared equivalent to 389.3173 m squared, divided by seconds squared, divided by kelvin, multiplied by our temperature T. And now we're just going to divide both sides of our equation by 389.3173 m squared, divided by seconds squared, divided by kelvin, so that it canceled out on the right hand side. So we're going to follow on the left hand side the same step. And this is going to leave us with Kelvin as our final unit for temperature. So we can say that our temperature is equal to 468. kelvin. And as the prompt states, we need our final answer to be in units of degrees Celsius. So we're going to recall that our kelvin temperature 468.673 kelvin is equivalent to our Celsius temperature added to 73.15 and so subtracting 2 73.15 from both sides, We're going to get our Celsius temperature equal to 194.9 degrees Celsius, which we can round to about 195 degrees Celsius for three sig figs. And this would be our final answer As the temperature of our sulfur dioxide gas molecules when they will have the same mean speed as our nitrogen dioxide gas molecules at 63°C. I hope everything I reviewed was clear. If you have any questions, please leave them down below what's highlighted in yellow is our final answer, and I'll see everyone in the next video.