An ideal gas is taken through the three processes (A →B, B →C, and C →A) shown in the drawing. I

An ideal gas is taken through the three processes (A →B, B...

An ideal gas is taken through the three processes $(A \rightarrow B, B \rightarrow C$, and $C \rightarrow A)$ shown in the drawing. In general, for each process the internal energy $U$ of the gas can change, because heat can be added to or removed from the gas and work $W$ can be done by the gas or on the gas. (a) For the process $A \rightarrow B$, is work done by the gas or on the gas? Why or why not? (b) For the process $B \rightarrow C$. suppose that the change $\Delta U$ in the internal energy of the gas and the work $W$ are known. Is it then possible to determine the heat $Q$ that has been added to or removed from the gas? If so, explain how this could be done. (c) For the process $C \rightarrow A$, is it possible to find the change in the internal energy of the gas if the change in the internal energies for the processes $A \rightarrow B$ and $B \rightarrow C$ are known? If so, specify how this could be done. For the three processes shown in the drawing, fill in the five missing entries in the following table. $$ \begin{array}{|l|c|c|c|} \hline \text { Process } & \Delta U & Q & W \\ \hline A \rightarrow B & (b) & +561 \mathrm{~J} & (\mathrm{a}) \\ \hline \mathrm{B} \rightarrow \mathrm{C} & +4303 \mathrm{~J} & (\mathrm{c}) & +2867 \mathrm{~J} \\ \hline \mathrm{C} \rightarrow \mathrm{A} & (\mathrm{d}) & (\mathrm{c}) & -3740 \mathrm{~J} \\ \hline \end{array} $$

An ideal gas is taken through the three processes $(A \rightarrow B, B \rightarrow C$, and $C \rightarrow A)$ shown in the drawing. In general, for each process the internal energy $U$ of the gas can change, because heat can be added to or removed from the gas and work $W$ can be done by the gas or on the gas. (a) For the process $A \rightarrow B$, is work done by the gas or on the gas? Why or why not? (b) For the process $B \rightarrow C$. suppose that the change $\Delta U$ in the internal energy of the gas and the work $W$ are known. Is it then possible to determine the heat $Q$ that has been added to or removed from the gas? If so, explain how this could be done. (c) For the process $C \rightarrow A$, is it possible to find the change in the internal energy of the gas if the change in the internal energies for the processes $A \rightarrow B$ and $B \rightarrow C$ are known? If so, specify how this could be done.
For the three processes shown in the drawing, fill in the five missing entries in the following table.
$$
\begin{array}{|l|c|c|c|}
\hline \text { Process } & \Delta U & Q & W \\
\hline A \rightarrow B & (b) & +561 \mathrm{~J} & (\mathrm{a}) \\
\hline \mathrm{B} \rightarrow \mathrm{C} & +4303 \mathrm{~J} & (\mathrm{c}) & +2867 \mathrm{~J} \\
\hline \mathrm{C} \rightarrow \mathrm{A} & (\mathrm{d}) & (\mathrm{c}) & -3740 \mathrm{~J} \\
\hline
\end{array}
$$

Key Concepts

First Law of Thermodynamics

This principle states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system (?U = Q - W). It provides the foundational energy conservation relationship for any process, allowing one to relate changes in heat, work, and internal energy.

Work Done by or on the Gas

Understanding the sign convention for work is key; work done by the gas is typically considered positive because the gas is expending energy, while work done on the gas is negative since energy is being added to the gas. This concept helps in analyzing energy transfers during expansion or compression processes.

State Functions

Internal energy is an example of a state function, meaning its change depends only on the initial and final states of the system rather than the path taken. This property allows one to relate the change in internal energy in cyclic processes or across multiple steps, even when the individual path contributions (heat and work) are different.

Path Dependence of Heat and Work

Unlike state functions, heat and work are path-dependent quantities. Their values depend on how the system transitions from one state to another, meaning that knowing the end states alone is not enough to determine the amount of heat transferred or work done without additional process information.

Cyclic Process Analysis

In any closed cycle, the net change in internal energy is zero because the system returns to its initial state. This concept allows one to use the sum of the energy transfers in individual processes to check for consistency and to calculate missing quantities, ensuring energy conservation over the entire cycle.

Transcript

00:02 In this problem, we have to analyze this particular pv diagram with this process from a to b, c, back to a.

00:13 And they ask us certain first conceptual questions.

00:17 The first one is, what is the work done from a to b? well, the work done from a to b is equal to zero.

00:25 There's no volume change.

00:31 From a graphical standpoint, there's no area under the curve, which gives us the work done.

00:36 In a certain process.

00:39 Just like if you have no displacement, there's no work in mechanics.

00:45 The next thing it asked, conceptually, is if you know the change in internal energy as you go from b to c, and you know the work as you go from b to c or you can determine that from a graph, could you find the heat, you know, from b to c, you know, that enters or leaves? and the answer is yes, and that's from the first law.

01:08 This is the first law, delta you, the change in internal energy for bc, process bc is qbc minus w.

01:18 Bc, that's the first law.

01:21 And we can just solve for qbc, delta ubc plus wbc.

01:35 The next question is, can you find the work done, or not the internal energy, i should say, change as you go from c to a, if you know the internal energy changes for a to b and b to c.

01:50 And the answer is yes, but this is a little more subtle.

01:55 First thing you got to remember is u is a state function.

01:59 Internal energy is a state function.

02:00 Now, what does that mean? that just means it does not matter how you got there.

02:09 It's the particular state.

02:11 You've got a certain situation.

02:13 You look at that situation and you have a certain value for the internal energy.

02:17 How the system got to that point is irrelevant.

02:22 Now, you might say, i've never heard of this before.

02:24 Well, yes, you have.

02:25 Gravitational potential energy, they don't talk about it.

02:28 Revitational potential energy, does it matter? how you go from, say, a height of two meters to a height of 10 meters? that potential energy at 10 meters is exactly the same if you go straight from 2 meters to 10 meters, or you zigzag, you know, in a weird path to go from 2 meters.

02:47 10 meters.

02:48 It is exactly the same.

02:49 So that's a state function also.

02:51 So with state functions, the change only depends on the begin and end points.

03:15 That's all really, it's really the same thing.

03:18 The pair, how you get there is out of the picture.

03:25 So with that said, we know we do the whole a, b, c back to a.

03:35 B, c, a...