A gas in a cylinder held at a constant pressure 1.80×10 5 Pa expands from a volume of 1.2 m 3 to 1.6 m 3 . The internal energy of the gas decreases from 4.40×10 5 J to 3×10 5 J. How much heat was transferred to the gas?

Hey, everybody. So let's get another practice problem here. So we have gas in a cylinder that's held at constant pressure. What happens is it expands from one volume to another, changes volume, and we're also told that the internal energy of the gas here decreases from this number to this number. And we wanna calculate not the change in the internal energy, but we wanna calculate how much heat is transferred to the gas. So basically, what we're looking for here is q 2. That's really our equation. However, we're still relating work and heat and internal energy, so we are still gonna use the first law of thermodynamics. So remember that's the change in internal energy of the gas equals heat added to the gas minus the work done by the gas. Okay? So this is what we're looking for here. So what I'm gonna do is I'm gonna start moving everything over to the other side. So I'm gonna move the W by to the other side, and what you're gonna get here is that the change in internal energy of the system plus the work done by the gas equals the heat added to the system. Right? So that's the equation. So what is the change in the internal energy of the gas? We're not told it directly, but we can figure it out because we're told that this internal energy decreases from this number to this number. So in other words, this is my energy initial and this is my energy final. Now remember whenever you have a delta, it always means final minus initial here. So what I can do is I can say, well, the change in the internal energy is just gonna be e final minus e initial. Right? That's always what delta means. So we have E final minus E initial and then plus with the work done by the gas. Remember, if it's at constant pressure it's just gonna be given by P delta V. So So I wanna do P times delta V. Now remember this is delta V here. The change in the volume was 1.1 or it's 1.2 to 1.6. So there's a couple of things you could do here. You could write out, you know, both of those numbers, or you could just look at this real quickly and say, well, delta v, if I'm going from 1 to 1.2 to 1.6, then it's just 0.4. That's one thing you can do here. So this is gonna be p times delta v, and this is gonna be q 2. So now we're just gonna plug in a bunch of numbers here. So we have the energy final, which is gonna be, let's see. This is gonna be just be careful with the signs here. It's gonna be 3 times 10 to the 5th minus 4.4 times 10 to the 5th. That's energy final minus energy initial. The plus, what you're gonna do is the pressure. The pressure I'm told is a constant 1.8 times 10 to the 5th, and then we've got the delta v, which remember is just 0 point 4. So that's gonna give you what the heat added to the gas is. Now just be really careful with the signs here because a lot of these problems are pretty straightforward. It's just the signs can trip you up and if you don't plug stuff in you're the wrong answer. But if you work this out carefully what you'll get is negative 6.8 times 10 to the 4th joules. Alright? So that's the final answer. If you look through your answer choices it's a. Alright. So that's it for this one. Let me know if you have any questions.

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A gas in a cylinder held at a constant pressure 1.80×105 Pa expands from a volume of 1.2 m3 to 1.6 m3. The internal energy of the gas decreases from 4.40×105 J to 3×105 J. How much heat was transferred to the gas?

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A gas in a cylinder held at a constant pressure 1.80×10⁵ Pa expands from a volume of 1.2 m³ to 1.6 m³. The internal energy of the gas decreases from 4.40×10⁵ J to 3×10⁵ J. How much heat was transferred to the gas?