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Ch 19: The First Law of Thermodynamics

Chapter 19, Problem 19

Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 27.0°C and the pressure is constant. As part of a machine design project, calculate the final temperature of the gas after it has done 2.40 * 10^3 J of work.

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Hey everyone welcome back in this problem. We have a silicate container containing 289 g of air case sealed with a piston. The gas inside the container is going to expand and it's going to do so under ice, a barrick conditions And produce 500 duels of work. Okay? What we want to do is we want to find the final temperature of the gas. We're told that the initial temperature is 290 Calvin and the molar mass of air is 28.9 g per mole. Alright, so, first, let's think about the conditions that were given. Okay, we're given ice of air conditions. Ice of air conditions. Tell us that we have constant pressure. Okay? And if we have constant pressure then we know we can write the work w is equal to p the pressure times delta V. Okay, Alright, now, we're talking about air. We can treat air as an ideal gas. We're doing that that we know we can use the ideal gas law. Now, let's recall the ideal gas law tells us that P delta V is equal to end our delta T. Okay, so our work, which is P delta V is going to be equal to N are times delta T. Okay, now, this lets us relate the work to the temperature. We want to find a final temperature. So, now we can relate the information we have and the information that we want. All right. So, in order to use this, we know the amount of work done. We know our as a gas constant. We know the initial temperature. We want to find the final temperature. Okay. So the only thing left to find right now is N. Now let's recall that N. Is equal to the mass M divided by the molar mass. Big M. In this case we have 289 g. The mass and the molar mass is 28.9 g per mole. Which is gonna give us an n. of 10 more. Alright, now that we have N. We can get back to our work equation. Okay. W. Is equal to N. R. Delta T. Which is going to be T final minus T. Initial. Hey, the work we're given is 500 jewels. The end we just found is 10 more R. Is the gas constant 8.314 jewels per mole Calvin. Okay. And then we have T. Final minus T. Initial which is 290 Calvin. Okay. And we're looking for T. Final. So now that's the only thing in this equation we don't have we can just rearrange and solve for T. F. Let's give ourselves a bit more room to work. All right, so we have 500 jewels on the left hand side. On the right hand side. We're gonna get 83.14. Okay? The unit of Mold will cancel. We're gonna be left with the unit of jewel per Calvin T. Final -290 Calvin Dividing by 83.14 jewel per Calvin. The unit of jewel will cancel. And we're going to be left with 6. Kelvin is equal to T Final - Kelvin. Which gives us a final temperature t. of 296.014 Calvin. Okay, so that is our final temperature of this ice a barrack Process. Okay. And that is going to correspond with answer. 296 Calvin. Thanks everyone for watching. I hope this video helped see you in the next one.
Related Practice
Textbook Question
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