A quantity of N2 occupies a volume of 1.0 L at 300 K and 1.0 atm. The gas expands to a volume of 3.0 L as the result of a change in both temperature and pressure. Find the density of the gas at these new conditions.

Video transcript

Hey everyone, Let's check out this problem at 306 kelvin and 2.1 80 M. An amount of C. O. Gas occupies a volume of 00.40 liters. What will be the density of the gas If the pressure of the gas is lowered and its temperature is raised until its volume is 3.8 L. So here we're solving for density and we know that our density equation is density equals pressure times molar mass over R. T. And here we're given two sets of information, we're given temperature, were given pressure and were given volume. And here we can say that we have D. One, P. One and T. One. Our molar mass is constant and so is our gas constant. And here we see that they give us volume but volume isn't in this equation. So a different equation that we can use to solve for density is really simple. It's D one times V. One equals D. Two times V. Two. And what we can do here is we can solve for D. One using the equation to the right and then plug it into this equation to solve for D. Two. So let's go ahead and solve for D. One first By plugging in what we know, so D one is equal to p times M over R. T. And we're told that the pressure is 2.1 A. T. M. And our molar mass of C. O. Gas using the periodic table. We have one carbon and one oxygen. That is 28.1 g per mole. And that's all over our T. R. R. Is r gas constant 0.8206 leader atmosphere over more Calvin. And our temperature is given at 306 Kelvin. So let's just make sure our units cancel our A. T. M's cancel our moles, cancel our kelvin's cancel. And we're left with units of grams per liter. Which is perfect because that is the units for density. So we'll plug this into our equation and we get 2.34 g per liter. So now we know D. One and we can rearrange this equation to solve for V. Two. Because we're solving for the new density, we're solving for detail, I'm sorry. We're solving for the new density. So let's go ahead and rearrange this equation and solve for that. And when we do that we get D. two is equal to D one times v. one over V. Two. So let's go ahead and plug everything in. We know that D. One what we just saw four is 2.34 g per liter & R. V one is 0.40 leaders And our V two Is 3.8 L. Both of our leaders cancel here. And we're going to be left with D two. And once we plug that into our calculator, we get 0.246 g per liter. And that's it. Everyone this is going to be the density of C. O. Gas. If the pressure of the gas is lowered and its temperature is raised until its volume is 3.8 liters. That's the end of this problem. I hope this was helpful.