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Calculating Work Done on Monoatomic Gas

Patrick Ford
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everybody. So, let's check out this problem here. So, we have the internal energy of a mono, atomic, ideal gas is given by this equation here, three halves. Nrt. Now, some of you may have already seen this equation, that's totally fine if you haven't. That's also fine. Just know that this is an equation for the internal energy where this end here is the number of moles in the gas, and t is the temperature. So, here's what's going on in this problem, We're gonna add 1300 joules of heat to some amount of gas, and its temperature is going to increase from 2 72 3 20. We want to calculate how much work was done by the gas. So, ultimately, we're trying to figure out what is w what's the w by the gas here. So, how do we start things off? Well, we're gonna have some heats, some works some internal energy. So, we're going to use the first law of thermodynamics. So, we're gonna write out this equation here. This change in internal energy of the gas equals the heat added to the gas minus the work done by the gas. Alright, so, if you want to calculate what this work done by the gas is, I'm gonna go ahead and rearrange some stuff. So, basically what I'm gonna do is I'm gonna move this to this side on the left side, and I'm gonna move the internal to the right side. What you end up with here, is that the work done by the gas is equal to the heat added to the gas minus the change in the internal energy of the gas. Okay, so that's really what's going on here, just moving some stuff around. So what is first the heat that's added to the gas here? So I'm gonna go over here and write out my variables. Well, if you look at the problem here we're gonna add 1300 joules of heat to the gas, so therefore Q two is equal to 1300. And we have what that variable is Now, what about this change in internal energy of the gas? That's really sort of what's going on in this problem? We're gonna have to go figure this out. Well remember that the change in the internal energy, the change in anything really is always equal to final minus initial. So it's the final minus initial here. So what are this the final and initial? What what happens is now we have a new equation to solve for this, it's just three halves N. R. T. So what's going on here is that the final minus E. Initial is really just equal to three halves N. R. T. Final minus three halves and R. T. Initial. Right so just final minus initial. The end is going to stay the same because it's just the amount of gas that I have. So that doesn't change. But what happens here in this problem is that we have some kind of increase in temperature. So this is my t. Initial and this is my t final. So that's what's going on. We have to use this equation to figure out what the change in internal energy is. And then once we figure this out, we can just plug that back into this equation and solve for the work done by the gas. Okay, so this is just gonna be three halves and we're gonna just go ahead and start plugging some stuff in. So this is gonna be 1.5, that's the number of moles that I have, This are constant, remember is just 8.314, you may remember that and then T final. So this is just going to be 320 Then we have -3/2, 1.5, 8.314 times to 70. Now you could have rearranged some stuff, you could have condensed this and shorten this equation, but I just plugged everything in, but which basically you're gonna get here, is that the change in the internal energy is equal to 935 jewels. So, remember this number here is what we plug into this in our original equation. So now, basically we're just gonna have to go and plug this last step in here, which is the work done by the gas is equal to Q to remember. This is just the 1300, that's the heat added minus the change in the internal energy here, which was just 935. So what you end up with here is that the work done by the gas is equal to 365 jewels. And that is the final answer here. Alright, so that's it for this one. Guys, let me know if you have any questions and I'll see you the next one.