A 100 cm³ box contains helium at a pressure of 2.0 atm and a temp...

Question

A 100 cm³ box contains helium at a pressure of 2.0 atm and a temperature of 100℃. It is placed in thermal contact with a 200 cm³ box containing argon at a pressure of 4.0 atm and a temperature of 400℃.

c. How much heat energy is transferred, and in which direction?

Accepted Answer

Hey, everyone. So this problem with dealing with heat transfer between gasses. Let's see what it we have two compressed gas cylinders, one containing neon and the other containing Xenon brought into contact with each other. The neon cylinder has a volume of 12 liters at temperature of 300 degrees Kelvin and a pressure of 2.5 atmospheres. The Xenon cylinder has a volume of 7.5 liters. A temperature of 400 degrees Kelvin and a pressure of two atmospheres. Assuming the two cylinders are thermally insulated from the surrounding environment. We're asked to calculate the amount of energy transferred between the two cylinders and the direction of that energy flow. Our multiple choice answers are the neon gains 120 jewels. While the Xenon loses 300 the Neon gains 400 Jews while the Xenon loses 400 the neon loses 550 Jews while the Xenon gains or the neon on. Sorry, the neon loses 650 Jews while the Xenon gains 350. So right off the bat, the problem tells us that the cylinders are thermally insulated, which means that we don't have any heat loss. So the amount that one gas gains is going to be equal to the amount that the other gas loses. And so we can eliminate answer choices A and D. And so now we can recalled that the internal energy of amount of atomic gas because given by the equation U equals three halves multiplied by N RT or N is equal to PV, divided by RT. And so we can find the initial energy of the neon gas and the Xenon gas 17 using the information from the prop. So, plugging this in, we have U equals three halves multiplied by PV, divided by RT multiplied by RT. So the RT is canceled and so you are initial, internal energy is simply three halves multiplied by PB. And so we're going to find the initial energy for the neon. So that will be three halves multiplied by the pressure is 2.5 atmospheres. We need to get that in standard units. So we're gonna multiply that by the conversion factor of 1. times 10 to the fifth pascals per atmosphere and then multiplied by V. The volume was given as liters. And to get that into standard units, we'll multiply that by 10 to the negative three m cubed per liter. And so we are left with an initial at R G of the neon of 4.6 times 10 to the third jules going to do the same thing for the xenon. So three halves multiplied by two atmospheres, use the same conversion factor. And that volume was given to us 7. liters. And so when we plug that into our calculator, we get 2.3 times 10 to the three tools we are looking for how much energy was gained or lost by the gas. So we have our initial energy. Now we need to find our final energy. And so we can recall that our equilibrium condition for mono atomic gasses is given by the equation Elon of the of one gas final is equal to the final of the second gas, which is equal to the total equilibrium. And so that can be rewritten as you hear energy U divided by the number of atoms. What happened? And the total is going to be the sum of the initial energies divided by the total uh molar mass. So um I'm sorry, I'm sorry, the total number of moles. So N of neon plus and for X, OK. So to solve this series of equations, we need to know N for xenon and neon. And we had written our equation for N already above. So we're going to solve that for both. And that was PV divided by RT. So P N V N divided by R T N. And so again, our pressure or neon, it's 2.5 atmospheres using that conversion factor to pass gals multiplied by 1. times 10 to the fifth Pass Vows per atom. Oh sorry. Here multiplied by our volume 12 A liters times 10 to the negative three m cubed per liter divided by R or gas constant. We can recall that's 8. for and our temperature of the neon was given and the problem and sweet like that and, and we get 1.22 months. And again, we can go on to do the same thing four power zenon. So P X V X divided by R T X. So we'll have two atmospheres multiplied by 1.13 times 10 to the fifth at scale for atmosphere multiplied by 7.5 liters times 10 to the negative three m cubed for leader all divided by same constant 8.314 joules per mole multiplied by the temperature of 400 K it's given in the problem. And so we plug that into our calculator and we get 0.457. No. OK. So we now have everything we need to solve for our final energies. So we'll take, but you final for neon, it's going to be equal to the sum of the initial energy for neon plus the initial energy for xenon divided by the moles of neon plus the moles of xenon all multiplied by the moles of neon. So we'll plug in our known values here so that would be 4.6 times 10 to the third tools plus 2.3 times 10 to the third Jews divided by 1. moles plus 0. mas multiplied by 1.22 moles. And that gives us 5.0 times 10 to the third jolt. And so we can do the same thing for Zon where the um first part of this equation doesn't change all that changes is instead of multiplying by the moles of neon, we're gonna multiply by the moles of xenon 0.457. And that gives us 1. times 10 to the three Jews. So the problem is asking for the heat gained or lost by the different gasses. And so our delta U for neon is going to be our final energy or five times 10 to the third jules minus our initial energy 4.6 times 10 to the third Jews or positive 400 jewels. And similarly, for Xenon delta U per Xenon is going to be our final 1.9 times 10 to the third Jes minus our initial 2.3 times 10 to the third jules equals negative 400 jules. And so that's the answer to this problem. Neon gains 400 Jews and Xenon loses 400 Jews. And that aligns with answer choice B. So that's all we have for this one. We'll see you in the next video

A 100 cm³ box contains helium at a pressure of 2.0 atm and a temperature of 100℃. It is placed in thermal contact with a 200 cm³ box containing argon at a pressure of 4.0 atm and a temperature of 400℃. c. How much heat energy is transferred, and in which direction?

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