A large cylindrical tank contains

Question

0.750

m³ of nitrogen gas at

27

°C and

7.50 times 10 cubed

Pa (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to

0.410

m³ and the temperature is increased to

157

°C?

Accepted Answer

Hey everyone in this problem. We have a laboratory technician that entrapped two mL of air at a pressure of 0.3 atmospheres inside the barrel of a syringe with a tightly closed hub. During an experiment, the increases the volume of the air to six mL by gradually moving the plunger back. The temperature of the air inside the barrel decreases from 25 degrees Celsius to 20 degrees Celsius. And were asked what is the pressures final value? Pf. Alright, so we're giving information about pressure, temperature and volume. And we're talking about air which we can treat as an ideal gas. So let's recall the ideal gas law which lets us relate the three pressure volume is equal to N. R T. Okay. And as the number of moles are as a gas constant. T. Is the temperature. Now we're going to have this for both. The initial state and the final state. Ok, so we're gonna have P one the initial volume. V one the initial or sorry, P one. The initial pressure V one, the initial volume is equal to N. R. T. One, and similarly, P F. V F is equal to N. R T F. Okay. And what you'll notice is that I haven't put a subscript on N or r. Okay. R is the gas constant? So it's constant, no matter what situation, where it's always the same and N is the number of moles and in our problem because the air is in this tightly closed syringe. There's not gonna be any change in the number of moles, there's nothing leaving or entering case we're gonna have a consistent number of moles. So let's go ahead and isolate for N. R. In both of these equations. So we have P one V. One over T one is equal to N. R. And similarly on the right wing at P F. V. F over T F is equal to N. R. Huh? Alright. So we have this expression on the left hand side equal to N. R. We have this expression on the right hand side equal to N. R. That means that those two expressions must also be equal. So this implies that we have P one V one Over T one equal PF V F over T F. So now we can relate the pressure, volume and temperature, both initially and finally. Okay, and what we're trying to find is a final pressure pf Okay, so we're trying to find this value. Let's go ahead and see what other information we know. Um and whether we can plug into this equation or whether we need to find some more information. So starting on the left hand side, P one. Okay, well we're told the initial pressure is 0.3 atmospheres. So P one is going to be 0.3 atmospheres. We want to convert this into pascal's. So we're gonna have 0.3 times 1. 13 times 10 to the five pascal's which is going to give us 30,390 pascal's so that's P One. What about V. One. Okay, well V one we're told is two mL the initial volume two mL. Now we want to convert this volume into meters. Cute. Okay, let's start by going from mill leaders to leaders and then we'll go from leaders two m cubed. Okay, so to go from milliliters to leaders, we're going to divide by 1000. This is going to give us 0.2 leaders and then to go from leaders two m cubed. We're going to divide by 1000 again and this is going to give us two times 10 to the negative six m cubed. Okay, so we were given milliliters, we divide by 1000 to get leaders. And then once we have our leaders, we divide by 1000 again to get meters cubed. All right Now, what about temperature? T1 were given the initial temperature we're told the temperature decreases from 25°C to 20. So the initial temperature is 25°C. Okay, we want to convert this to Calvin. So we're gonna take 25 and add 273.15 to put this into Calvin. It's gonna be 298.15 Calvin. Okay, so we know our P. One. V. One and T. One. Everything on the left hand side of our equation we know. Okay, let's move on to the final values. Gonna leave up our equation to give ourselves some more room to write. Okay so P. F. Again. This is our final pressure that we were looking to find the final volume. Okay we're told that the volume is increased to six mL. So our final volume is six mL. And again we want to convert this to meters cubed. So we divide by 1000 to get 0. liters and then we divide by 1000 again to get six times 10 to the negative six m cubed. Okay just like the one mL to liters divide by liters. two m cube divide by 1000. And what about our final temperature? We're told the temperature decreases from 25 degrees Celsius to 20 degrees Celsius. And so our final temperature is 20 degrees Celsius converting to Calvin again 20 plus 273. which gives us 293.15 kelvin. Okay so we have all the values we need we have P. One V. One and T. One. We have V. F. And T. F. The only thing we don't have in this equation is pf which we're looking to find. So we can plug in our values and sell for P. F. So plugging in our values we get 30, pascal's times two times 10 to the negative six m cubed, divided by 298.15 Calvin is equal to B. Two. Sorry not P. Two P. F. Is what we're calling it. The final pressure times six times 10 to the negative six m cubed. K. That final volume divided by the final temperature 293 by 15 Calvin. Hi. All right. Now when we do this division, okay, we're gonna multiply by 293.15 Calvin. We're going to divide by six times 10 to the negative six m cubed. So the units of meters cubed and Calvin will cancel. And we're gonna be left with the final pressure P. F. Of 9960. pascal. Okay. This is our final pressure. And if we go back up to our answer choices, okay. We see that we're gonna have answer choice. Okay so the pressure we found 9960. pascal's can be written approximately as 0. times 10 to the four pascal's. Thanks everyone for watching. I hope this video helped see you in the next one

Key Concepts

Ideal Gas Law

Charles's Law

Boyle's Law

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