What volume is occupied by 12.5 g of argon gas at a pressure of 1.05 atm and a temperature of 322 K? Would the volume be different if the sample were 12.5 g of helium (under identical conditions)?

Video transcript

hey everyone for this problem. We want to know what volume 12.5 g of argon gas occupies at a pressure of 1.2 to a T. M. and a temperature of 352 Kelvin. And if the sample were helium, would the volume be different under identical conditions? So here we're dealing with pressure, we're dealing with temperature and volume. So we can use the ideal gas law equation which is PV equals N. R. T. And here we're solving for volume. So let's go ahead and isolate that variable. We'll divide both sides of our equation by P. And we're left with volume equals N. R. T over P. So let's go ahead and plug in what we know for the first part of the equation which is dealing with argon gas. So our volume is equal to N. Which is our moles. And here they give us 12.5 g of argon gas. And we're going to convert that to moles of argon gas using its molecular weight. So in 12.5 g of argon we have one mole is equal to 39.95 g. And that's using our periodic table. Our units cancel And we're left with zero 313 malls of argon. So we'll plug that in and our our is our gas constant Which is 0.08206 leader A T. M over more Calvin. And our temperature is given at 352 Kelvin and this is all over? Our pressure Of 1.2 - eight p.m. So let's make sure our units cancel here are moles cancel Our ATMs cancel and our Kelvin's cancel. And we're left with the unit of leaders, which is perfect because we're looking for a volume. So let's go ahead and plug that into our equation our into our calculators. And when we do that we get our volume is equal to 7.40 leaders of argon gas. Now, it's also asking if the sample were 12.5 g of helium, would the volume be different under identical conditions? And so we're going to do this over. But instead of using our argon we're going to use helium. So we get volume is equal to N so our moles here is going to be 12.5 g of helium. And in one mall of helium we have 4.003 grams of helium. And we get our units cancel. We're left with 3.12 moles of helium. So let's plug that in. We'll get 3.12 moles of helium and everything else is going to be the same as above our gas constant, our temperature and our pressure. Okay, so our units cancel moles of helium, our ATMs cancel and our kelvin's cancel. And we're left with a volume of 73.9 leaders of helium. So if the sample were the same, the volumes would be different under identical conditions. We see that our volume of argon gas is 7.40 leaders, and our volume of helium is 73.9 L. And that's the end of this problem. I hope this was helpful.