Oxygen (O2) has a molar mass of 32.0 g>mol. What is (a) the average translational kinetic energy of an oxygen molecule at a temperature of 300 K;

Video transcript

Hey everyone welcome back in this problem. We are asked what is the mean translational kinetic energy of atmospheric nitrogen? We're told the more massive atmospheric nitrogen is 28. g per mole. And we're at a temperature of 298 Calvin. Alright, so let's recall the mean translational kinetic energy. K tr Can be written as 3/ times and are T. U. Okay, we know the temperature T and R. Is r gas constant. So we know that but we don't know N. Okay, we do know the molar mass but not but we know that we can relate end to the molar mass by writing N. Is equal to M divided by bigger. Okay, so the mass divided by the molar mass. So that means that we can write our translational kinetic energy as three halves mass divided by Mueller mass times. Rt Alright, so now we have the molar mass in there which we know but we have the mass as well. We don't know the mass. So we also need to figure out how we can determine the mass or manipulate this equation so that we don't need to know the mass. Now let's recall as well, we have a relationship between the molar mass and the mass where M big M. Mueller mass is equal to little M. The mass times avocados number and a Alright so in our equation we have little M divided by big M. Okay, so let's try to get little M divided by big M. Here as well. So if we take little M divided by big M. Okay we're gonna have to divide avocados number until left hand side. We're gonna get one divided by I've got your number. Okay. And so we have this expression the ratio of the mass to the molar mass is going to be equal to one divided by salgado's number. Well now we can replace that in our equation. So now we have the mean translational kinetic energy is equal to three. RT. They divided by two and java graduates number. And now our main translational kinetic energy only depends on the temperature. Okay. R and N A R. Constance. It only depends on the temperature, we know the temperature so we can find our translational kinetic energy. So substituting the values we know we have three times are which is 8.314. We have the temperature we're told is 298 Calvin Divided by two times avocados number 6.02. 2 Times 10 to Exponent 23. Okay. And this is going to give us a translational kinetic energy with 6.17 times 10 to the negative jewels. Okay. And so that is the mean translational kinetic energy for atmospheric nitrogen at the temperature of 298 Calvin K. We have answered D 6.17 times 10 to the negative jewels. That's it for this video. Thanks everyone for watching. See you in the next one