Calculate the internal energy change for each of the...

Calculate the internal energy change for each of the following. a. One hundred $(100 .)$ joules of work is required to compress a gas. At the same time, the gas releases 23 $\mathrm{J}$ of heat. b. A piston is compressed from a volume of 8.30 $\mathrm{L}$ to 2.80 $\mathrm{L}$ against a constant pressure of 1.90 $\mathrm{atm}$ . In the process, there is a heat gain by the system of 350. J. c. A piston expands against 1.00 atm of pressure from 11.2 $\mathrm{L}$ to 29.1 $\mathrm{L}$ . In the process, 1037 $\mathrm{J}$ of heat is absorbed.

Transcript

00:01 So for all these questions, we're trying to find the internal energy.

00:03 So for question a, we're told that we have 100 joules of work.

00:09 Then we're being told that work is required to compress the gas.

00:14 So the system needs the work, which means that the work is being done by the surroundings on the system.

00:22 And when work is being done by the surroundings onto the system, it's going to be positive.

00:27 Then we were told that we, we're told that we, we have 23 joules of heat, which is our q, and then we're told that the gas releases the heat.

00:38 So if the gases are system and it's releasing the heat into the surroundings, it's losing that heat, so it's going to have a negative.

00:46 Then we use our equation where the change in internal energy is equal to q the heat plus w the work.

00:54 So our change in internal energy is going to be equal to q, which was negative 23 joules, plus our work, which was a positive 100 joules.

01:04 Put that into your calculator, and you should get 77 joules.

01:10 Now we'll go to part b.

01:12 We're told that a piston is compressing in volume, where our initial volume is 8 .3 liters, and our final volume is 2 .8 liters.

01:27 Our pressure is constant at 1 .9 atmospheres.

01:34 And that the heat involved is 350 joules.

01:39 And we're told that the heat is being gained by the system.

01:43 So if the system is gaining the heat, it is going to be a positive number.

01:49 So we have our cue for the equation, but now we need our w.

01:52 So the other equation that we're going to use is our work is equal to negative the pressure times the change in volume.

02:00 So our work is going to be equal to negative the pressure, 1 .90 atmosophobic.

02:05 Times the change in volume, which is the final volume, 2 .8 liters, minus the initial volume, 8 .3 liters.

02:16 Type that into your calculator, and you should get positive 10 .45 liters per atmospheres.

02:25 Now, before we can put it into this equation, we need to convert these units into joules.

02:31 So we have 10 .45 liters times atmosphere.

02:35 So we can take it.

02:36 Take it time the conversion rate of 101 .3 joules for every one liters times atmospheres.

02:43 The liters atmospheres units will cancel 10 .455 times 101 .3...

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