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.

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.

Key Concepts

First Law of Thermodynamics

The first law of thermodynamics is the principle of energy conservation applied to thermodynamic processes. It states that the change in the internal energy of a system, ?U, is equal to the sum of the heat transferred into the system, Q, and the work done on the system, W. This relation, expressed as ?U = Q + W, is fundamental for calculating energy changes during processes such as compression or expansion.

Heat Transfer

Heat transfer refers to the energy exchanged due to a temperature difference between the system and its surroundings. In thermodynamic calculations, the sign of the heat is important: positive Q indicates heat absorbed (gained) by the system, while negative Q indicates heat released (lost) by the system. The concept is crucial for determining the direction of energy flow in a process.

Work Done on a System

Work done on or by the system is another key component in thermodynamic processes. The sign of work in the energy balance depends on the convention used; here, work done on the system is taken as positive, meaning that when the system is compressed (work is done on it), W is positive, and when the system expands (it does work on the surroundings), W is negative. Understanding this sign convention is pivotal for correctly applying the first law of thermodynamics.

Unit Conversions in Thermodynamics

In thermodynamics, it is often necessary to convert between different units of energy. For example, work calculated from pressure and volume changes may be expressed in liter-atmospheres (L?atm), which can be converted to joules using the relation 1 L?atm ? 101.3 J. This conversion ensures consistency of units when applying the first law of thermodynamics to compute internal energy changes.

Transcript

00:01 In this video, we'll determine delta e for three separate problems for this particular question.

00:09 So starting with part a, we are given that 100 joules worth of work is put into the system and that we have 23 joules worth of heat going out.

00:24 And i put it off to the side that since the gas is being compressed, we know that work is going in.

00:31 So we know work is going to have a positive sign.

00:35 And with heat going out, the q, 23 joules for q will have a negative sign.

00:42 So we'll go ahead and run through the numbers real quick on part a.

00:47 We know delta e is the value of q plus w, where q is heat in or heat out and work is work in or work out.

00:59 So for q, we know we have 23 joules out.

01:07 So that would be negative 23 joules.

01:14 And then the work is going into the system.

01:20 So we'll just go ahead and net that out to 77 joules.

01:26 That would be positive.

01:28 So that would be part a.

01:30 Part b, we're given some information about a similar system.

01:36 We're given that work is going to, in due to compression.

01:42 Compression is a key term indicating that work is going in.

01:46 And then heat is also going in because they indicate it's a heat gain.

01:50 So we know heat is going into the system.

01:53 So after the side, i'm showing the gas chamber with q going in as well as the work going in.

02:00 So if we go ahead and run the numbers on that, we'll go ahead and do this.

02:08 Delta e equals again q plus okay.

02:20 Now, we do have to use before we actually, we can go ahead and add the 350 joules.

02:29 That's the heat going into the system right here.

02:32 That's that right there.

02:40 That's the cue.

02:41 Now, we need to find the work going in.

02:44 So we have to go back to the problem.

02:46 They say we're working against 1 .9 atmospheres of pressure.

02:53 And we're being compressed from 8 .3 liters down to 2 .8 liters.

02:58 So 2 .8 liters is our final volume, and 8 .3 is our initial volume.

03:05 So we're actually working off negative p delta v, where delta v is final volume minus initial volume.

03:14 So we'll go ahead and put that negative sign out there.

03:16 So we're going to have, if we run the math through here, we're going to have 10 .45 liter atmospheres.

03:35 Okay.

03:36 Now we do need to put that in terms of jewels...

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