A 334-mL cylinder for use in chemistry lectures contains
5.225 g ...

Question

A 334-mL cylinder for use in chemistry lectures contains
5.225 g of helium at 23 °C. How many grams of helium must
be released to reduce the pressure to 7.60 MPa assuming ideal
gas behavior?

Accepted Answer

hi everyone for this problem. It reads at 26 degrees Celsius a millimeter gas tank in a laboratory contains 43. g of argon calculate the mass of argon n grams that must be released in order to reduce the pressure to 4.66 mega pascal's assume that argon gas behi ideally. Okay, so what we want to do here is calculate the mass of argon specifically in grams. And because we're dealing with a gas here, we want to use the ideal gas law to solve this problem, the ideal gas law is pressure times volume equals moles times r gas constant times temperature. Okay, so to calculate the mass, we're going to want to first know how many moles of argon we have and that is represented by our variable end. So we want to rearrange this equation so that it's solving for N. So we'll do that by dividing both sides by R. T. And we'll be left with N. Which is our number of moles is equal to pressure, times volume over r. Gas constant times temperature. Okay, so let's look at what we have. So our gas constant R is equal to 0. Leaders times a tm over mold times kelvin. So we want to make sure that all of the units that were using match these units here. So that means for pressure. It needs to be an A T. M. R. Volume needs to be in leaders and our temperature needs to be in kelvin. So let's start off with our pressure. We're told that our pressure is equal to 4.66 mega pascal's. So we need to go from mega pascal's to A T. M. Okay? So let's go ahead and get started. And one mega pascal there is 10 to the sixth pascal's. So looking at our units are mega pascal's cancel. And we're left with pascal's. Now we need to go from pascal's to A T. M. And one A. T. M. There is 10, 100. No excuse me. 101, pascal's. Okay, so looking at our units here, our Pascal's cancel. And we're left with ATM which is the unit that we want. So this gives us 45.99 A. T. M. Perfect. Now our volume needs to be in leaders and in the problem we're given the volume to be 556 ml. Okay so we want to go from ml to leaders. So in one leader there is ml. So we're going to divide five. We're going to divide 556 by 1000 C here are units cancel and we're left with leaders and we get 0.556 liters. Great. Lastly temperature our temperature needs to be in Calvin. But in the problem we're told it's 26 degrees Celsius. So we need to go from degrees Celsius to Calvin. And the way that we do that when we're in Celsius. And we want to go to kelvin is we add 273.15. So that will give us a temperature of 299. kelvin. So now we have everything we need to plug into our equation to solve for moles. Okay, So let's go ahead and do that. So we get and is equal to our pressure, which is 45.99 A. T. M. Times are volume 0. liters divided by r r gas constant 0.8206 leaders times A T. M. Over more times Calvin. And this is multiplied by temperature 299. Kelvin. Let's make that a little clearer. Okay? Alright, so we want to make sure our units cancel properly here. So let's go ahead and do that. We should be left with just moles. Okay? Because that's what we're solving for. C R A T m's cancel. Our leaders cancel, kelvin's cancel and we're left with moles. Okay? Which is what we're looking for. So let's go ahead and plug this into our calculator. And when we do we get 1. 164 moles of Argon. The problem asks us to calculate the mass of argon and grams. So we have it in moves. So we need to go from moles to grams. And the way we go from moles to grams is by using molar mass. So we need to pull out our periodic tables. Okay, so let's go ahead and write out what we're starting with. We have 1. moles, 0.1 point 04164 moles of argon. And we want to go to grams. So and one more of argon. Our Molar mass is 39. g of argon. Okay, so our moles cancel and we're left with grams. So we get 41.6 g of argon. Now looking at our problem, It's asking us to calculate the mass of argon that must be released. Okay, so that means we need to see what the difference is between what we started with and what the masses. Okay, so we started with 43.6g. Okay, so we have 43.6 g minus what we just saw. 4 41.6 g. This gives us two g of argon. That's how much was released in order to reduce the pressure to 4.66 mega pascal's. So our final answer here is going to be this two g of argon. That's the end of this problem. I hope this was helpful

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