Additional Problems Suppose 2.00 mol of a diatomic gas is...

Additional Problems Suppose 2.00 mol of a diatomic gas is taken reversibly around the cycle shown in the $T$ - $S$ diagram of Fig. $20-35,$ where $S_{1}=6.00 \mathrm{J} / \mathrm{K}$ and $S_{2}=8.00 \mathrm{J} / \mathrm{K}$ The molecules do not rotate or oscillate. What is the energy transferred as heat $Q$ for (a) path $1 \rightarrow 2,(\mathrm{b})$ path $2 \rightarrow 3,$ and (c) the full cycle? (d) What is the work $W$ for the isothermal process? The volume $V_{1}$ in state 1 is 0.200 $\mathrm{m}^{3} .$ What is the volume in (e) state 2 and (f) state 3 ? What is the change $\Delta E_{\text { int }}$ for (g) path $1 \rightarrow 2,$ (h) path $2 \rightarrow 3,$ and (i) the full cycle? (Hint: (h) can be done with one or two lines of calculation using Module 19-7 or with a page of calculation using Module 19-9.) (j) What is the work $W$ for the adiabatic process?

Additional Problems
Suppose 2.00 mol of a diatomic gas is taken reversibly around the cycle shown in the $T$ - $S$ diagram of Fig. $20-35,$ where $S_{1}=6.00 \mathrm{J} / \mathrm{K}$ and $S_{2}=8.00 \mathrm{J} / \mathrm{K}$ The molecules do not rotate or oscillate. What is the energy transferred as heat $Q$ for (a) path $1 \rightarrow 2,(\mathrm{b})$ path $2 \rightarrow 3,$ and (c) the full cycle? (d) What is the work $W$ for the isothermal process? The volume $V_{1}$ in state 1 is 0.200 $\mathrm{m}^{3} .$ What is the volume in (e) state 2 and (f) state 3 ? What is the change $\Delta E_{\text { int }}$ for (g) path $1 \rightarrow 2,$ (h) path $2 \rightarrow 3,$ and (i) the full cycle? (Hint: (h) can be done with one or two lines of calculation using Module 19-7 or with a page of calculation using Module 19-9.) (j) What is the work $W$ for the adiabatic process?

Key Concepts

Isothermal Process

An isothermal process occurs at a constant temperature, meaning that any heat transfer to or from the system must be compensated by work to maintain that constant temperature. This process is fundamental in understanding energy exchange mechanisms in cycles where temperature remains steady.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings, meaning that any change in a system’s internal energy is solely due to work done. This concept is crucial for assessing processes where the rapid change in state prevents heat exchange, thereby simplifying the energy balance equations.

Entropy Change

Entropy is a measure of the disorder or randomness in a system, and its change during a reversible process is directly linked to the heat transfer by the relation dS = dQ_rev/T. This concept is key in determining the feasibility and efficiency of processes and in analysing the direction of spontaneous processes.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, is fundamental in connecting the macroscopic variables of pressure, volume, and temperature for an ideal gas. It underpins the calculations of state variables during thermodynamic processes and is used to relate changes in these variables as the system evolves.

Internal Energy of Ideal Gases

For ideal gases, the internal energy depends solely on temperature and the molecular degrees of freedom. Understanding this relationship is crucial because it allows for the calculation of internal energy changes in processes, even when the system undergoes various thermodynamic pathways.

First Law of Thermodynamics

The first law of thermodynamics states that the change in the internal energy of a system is equal to the net heat added to the system minus the work done by the system. This principle provides the foundational relationship linking heat, work, and internal energy, which is essential for analyzing any thermodynamic process.

Temperature?Entropy (T?S) Diagram

The temperature-entropy diagram is a graphical tool that maps the state of a system with temperature on one axis and entropy on the other. The area under a process curve in a T?S diagram represents the heat transferred during a reversible process, making it a valuable visualization for analyzing thermal cycles and energy exchanges.

Reversible Process

A reversible process is a theoretical construct in which the system changes state in such a way that the process can be reversed by an infinitesimal modification of conditions, ensuring that the system remains in thermodynamic equilibrium at all points. This concept is central for establishing maximum efficiency limits and for relating heat transfer to changes in state functions.

Transcript

00:02 So here, essentially from equation 21 for part a, we can find the, we can essentially infer that here the heat transfer would be equal to the integral of the temperature times ds.

00:20 In this case, this corresponds to the area under the curve in a ts diagram.

00:26 So here, this would simply be equal to area under curve in t temperature versus entropy graph.

00:40 And if this is the case, the area is simply the area of a rectangle.

00:45 So the heat transfer from 1 to 2 would be equal to 350 times 2 .0.

00:55 This is equaling 700 joules.

00:58 For part b, we know that here from 2 to 3 there is no area under the curve.

01:13 So we may say that q from 2 to 3 would equal 0 joules.

01:19 For part c, we can say that the cycle, for the entire cycle, the net heat should be the area inside the figure.

01:29 Here, if it's the entire area, we can say that simply q net would be the area of the triangle.

01:37 So this would be one half times the base times the height or one half times 2 .00 times 50.

01:43 And this is equaling 50 joules.

01:47 At this point, we can say for part d, we are dealing with an ideal gas, ideal gas isothermal process.

02:02 If this is the case, the change in internal energy is going to be equal zero joules.

02:09 Therefore, work sub 1 to 2 would be equal to q sub 1 to 2, and this would be equal to, again, 700 joules.

02:24 So that would be our answer for part d, answer for part c, part b, part a.

02:30 Let's move on to d.

02:32 It's an e rather.

02:32 So for e we're going to have to use equation 1914.

02:39 And at this point, we'll use for isothermal work.

02:48 So if we're using it for isothermal work, we know that then the work from 1 to 2 is going to be equal to nrt natural log of v .2 divided by v.

02:59 Sub 1.

02:59 And so if we know that t is equaling 350 kelvin and we know that v sub 1 is equaling 0 .2 -0 cubic meter, we can then say that v.

03:16 2 is going to simply equal v.

03:18 Sub 1, e -xp.

03:20 When i say the exponential function, the exponential function is simply used in order to write it a little bit neater, but what it essentially means is that anything within the parentheses, is the power with a base of e.

03:33 So we can say the exponential of w over nrt.

03:39 So what this means essentially is v .1 times e to the work over nrt power.

03:46 And then we can solve.

03:48 So 0 .200 times e to the 0 .12 power...

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