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# 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$^\circ$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 $\times$ 10$^3$ J of work.

## $$=75.1^{\circ} \mathrm{C}$$

#### Topics

The First Law of Thermodynamics

### Discussion

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##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

##### Farnaz M.

Simon Fraser University

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### Video Transcript

So the question states that six moles of an ideal gas are used in a cylinder fitted at one and with the movable piston. The initial temperature of the gases twenty seven degrees C and the pressure is constant was a part of a machine designed project. Calculate the final temperature of the gas after it is done two point four times ten other three jewels of work. So we just want to start out by writing out what we're given. So that's on equals six Smalls and that the work is equal. Tio two point four times ten to the three jewels and the initial temperature ISS twenty seven degree sea. And our goal is to get the final temperature. It's all right equals question Mark create. So just to give an overview of what we're going to do to solve the problem, we're going to use thie. First, use the definition of work being thie integral of P Delta V to obtain the quantity p Delta V. And then we're going to use that I don't gas law to really people to be Teo the change in temperature. So starting out that we can say that in the case of a constant pressure that the work is equal to p. Still to be so the integral reduces to that. And I'm not gonna worry about Sign. So it'Ll be implied. It's a magnitude, and we can stop this equal, Tio. And actually, I'm going to use the same symbol to be consistent. And you are, uh, Dr Tear to find the change in temperature so we can alter Brickley manipulate this toe to say the Delta T is equal to p doll to be divided by on our way. Um and then we know that people to be witches. The work is two point four times ten to the three tools and on DH six moles. And then we need to use the ideal gas constant eight point three one Jules promote Calvin. It's a little sloppy. Um, way confined. The daughter tea is equal. Teo, sixty, works his mean thirty five point one Calvin. Yeah, if you find out into a calculator and I noticed the unit's workout. Um, if we cancel those jewels and those moles put the comments on the top and just does a note since Degrees Celsius is related, Teo degree cabin by an additive to seventy three. So you just ad to seventy three to the degree Celsius to get the T final than a change and temperature of thirty five point one. Calvin is actually the same as a change in temperature of thirty five point one degrees C. So to get the final temperature, we can just say that t final equals T initial, which is twenty seven degrees plus thirty five. I'm just going to cut the point one because you only have to say things for the other temperature because you see degree c on, then that's equal to sixty two degrees C. So that should be you are final temperature.

University of Washington

#### Topics

The First Law of Thermodynamics

##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

##### Farnaz M.

Simon Fraser University

Lectures

Join Bootcamp