🚨 Hurry, space in our FREE summer bootcamps is running out. 🚨Claim your spot here.



Numerade Educator



Problem 10 Easy Difficulty

Five moles of an ideal monatomic gas with an initial temperature of 127$^\circ$C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?


$$=117^{\circ} \mathrm{C}$$


You must be signed in to discuss.

Video Transcript

So the question reads that five moles of an ideal Mon atomic gas, with an initial temperature of one twenty seven degrees Celsius, expands and in the process absorbed fifteen hundred jewels of heat and does twenty one hundred jewels of work. What is the final temperature of the gas? So let's start out by writing on our Gibbons. Then we can discuss which formulas to use. Do you have an end equals five Moors and we have t initial is one twenty groups one twenty. Go ahead and do that. Oh, one twenty seven degrees C. And we know that Q Hugh equals fifteen hundred drools. And we know that work is twenty one hundred joules, and our goal is to get the final temperature. Um so first we want to start out with the equation that the internal energy you is equal to or the changing internal energy is equal to the heat input to the system minus the work done by the system. And we know that the internal energy is always equal to three house and are the local head of adult for a change. And then if the numbers days of saying that's equal to and our delta T because generally it's equal to three house and our teeth. So the changes three house and our Delta team, and then we know that fifteen hundred rules of heat are added, and then we know that worked. The what is that? The gas does twenty one hundred jewels of work. Um and so looks like we have all the ingredients that we need to solve for the final temperature. So we can say just algebraic curly and manipulate those, um, and get out ot is fifteen hundred Charles, my as twenty one hundred jewels. We'LL divide it by three house and our, um some kind of mixing using symbols and numbers. But hopefully it's still clear what I'm doing. So if you play that into a calculator, um, you got that? The daughter tea is minus fourteen degrees C. And so therefore, if you get your to get the final temperature, yeah, you get you to your initial minus fourteen. And if you take the initial temperature of one twenty seven and you subtract fourteen, then you got one thirteen degrees C on DH, then, just to know we didn't have to convert to Calvin because Calvin is additive with Celsius. So to get from coven of Celsius, you just add two seventy three. So if you're only right out temperature differences, you can use Celsius and Calvin interchangeably. Um, yeah. So that's the problem.

University of Washington
Top Physics 101 Mechanics Educators
Elyse G.

Cornell University

Andy C.

University of Michigan - Ann Arbor

Jared E.

University of Winnipeg

Meghan M.

McMaster University