The adult blue whale has a lung capacity of 5.0×103 L. Calculate the mass of air (assume an average molar mass of 28.98 g/mol) contained in an adult blue whale’s lungs at 0.0°C and 1.00 atm, assuming the air behaves ideally.

Video transcript

Hello everyone in this video. We're going to be dealing with the ideal gas law equation. So first let's start off by writing out what given information we have. So information we're given right off the bat. I can see that we're given a temperature At 27.0° C reading a lot further we get a volume Of 9 85 ml. Then We also have a pressure of 0.5 P. S. I. So the ideal gas law equation is also known as piven Art because it's PV equals N R. T. So from the given information I can see that I have a temperature, I have a volume and I have a pressure. So we have our P value V value and T value are something that is always constant. And it comes along with our ideal guest law equation because it's the ideal gas law constant. So capital R is always going to equal 0.08206. The units are a T. M times leaders per kelvin time small and because this is a constant, the units will never change and the value will never change. And because of that, any information that we're going to add into our PV equals NRT equation. We want the units to match for the constant as well. And you can see here that we want to find our end value which is our moles. But the problem asked for the mass. So first we can find the moles from this equation that we do to Misha analysis and get the gramps and mass. Alright. So what we're trying to find here from this problem is going to be the very end of the mass, nitrogen and grounds. Alright perfect. So because we're solving for just the end value, I want to go and manipulate my equation, divide each side by R. T. And why I want to do that is because I'm isolating my end. So you see here that will cancel out. I said like me with just the end I'll go ahead and rewrite that equation. So my new equation that I'll be utilizing is N. Equals P. V. All over our tea. So plugging into my values. Let's see first we have the units of Celsius for temperature but we want kelvin's we have the units of male leaders in volume. But we want leaders and we have P. S. I. For pressure but we want A. T. M. So let's go ahead and do some providence first. Before we can go ahead and plug in any information into our ideal gas law equation. So first starting off with our temperature We are given 27.0°C. How we can convert that Calvin's is by adding 273 .15. So plugging that into my calculator I would get 300.15 Kelvin's. Next we have the volume It's given to us in ml. So 985 ml. I'm gonna go ahead and convert that into leaders. So we have 1000 ml for every leader. You can see here that the ml will cancel out, giving us a value or the answer in leaders. Bring that into my calculator. I'll get 0.985 leaders. Lastly, We have the 96.5 P. S. A. But we want a. T. M. So the conversion for that is every A. T. M. We have 14.69 P. S. I. And canceling units out. We can see we're left with A T. M. Which is what we want. Putting that into my calculator. I'll get the value of 6.5 A. T. M. Alright, so let's go ahead and now use this. So I'll have to scroll down a little bit. I'll have the moles equaling to the pressure. So that's going to be 6.569 A. T. M. Multiplied by the volume of 0.985 L. That's all going to be divided By the gas law constant. So 0.08206 units of ATM times. Leader per kelvin time is small and multiplied by the temperature. That's going to be 300.15 Kelvin's. So canceling out some units, you can see the A T M's will cancel. Leader will cancel and kelvin will cancel. So that's why we need to match up everything with the gas law constant units. So playing that, putting everything into my calculator, I'll get the value for the most of 0.26 one law of into gas. So like I said before, now that we have the moles, we want to convert that into the mass, which is in grams. So scrolling down some more. We're starting off with the 0. moles are end to gas and we know that each atom of our nitrogen That has a molar mass of 14.1. So 14.1. Let me just write that off to the side 14. times two. Because we have to Adams of nitrogen that gives us a value 28.02 g per. And that would be my competitive vector. So on the bottom, I'll have one more and to Now on top 28.02 g of the end too. See here that the moles cancel out leaving us with just the grams of our end too. Putting that into my calculator, I would get 7.36 g and two. And that's going to be my final answer for this problem. Thank you everyone for watching