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Ch 18: Thermal Properties of Matter

Chapter 18, Problem 18

(a) Compute the specific heat at constant volume of nitrogen (N2) gas, and compare it with the specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.

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Welcome back everybody. We are making some observations about oxygen. Now for oxygen, we know that the moller mass is 32g per mole. Now. Just to work with this, I'm just gonna go ahead and convert this two kg. Right there are 1000 g in one kg. These units are going to cancel out giving us that our molar mass of oxygen is 32 times 10 to the negative third kilograms per mole. And we are tasked with finding two different things here. We are tasked with finding first off the specific heat of oxygen and to how the specific heat of oxygen compares with the specific heat of water. And the way we are going to compare the two is find some factor to multiply by the specific heat of oxygen. So let's go ahead and start out with part One here, finding the specific heat of oxygen. Well, we have a formula that states that the molar heat capacity is equal to the molar mass times the specific heat. In this case, specific heat of oxygen. I'm gonna divide both sides by molar mass so that I can isolate the specific heat which yields that. Our specific heat of oxygen is just equal to our molar heat capacity divided by our molar mass Were given both of these things. So let me just go ahead and plug them in and then we will find our specific heat of oxygen. So our molar heat capacity is 20.85 divided by 32 times 10 to the negative third, giving us a specific heat for oxygen of joules per kilogram Calvin. Wonderful. Moving on to part two. Here, we need to compare the specific heat of water to the specific heat of oxygen via some constant. So what I'm gonna do is I'm actually gonna divide by the specific heat of oxygen on both sides. To isolate this constant term so that we can find it. So X. Is equal to specific heat of water divided by the specific heat of oxygen. You can look at the specific heat of water in your tables and it comes out to be 4180 jewels per kilogram kelvin. And we just found the specific heat of oxygen to be joules per kilogram Kelvin units are going to cancel out. And we are left with the X. is equal to 6.42. So this means that the specific heat of water is approximately six times greater than that of the specific heat of oxygen. So we found the specific heat of oxygen. We compared it to the specific heat of water which gives us our final answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one
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