Suppose 2.00 mol of a diatomic gas is taken reversibly...

Suppose 2.00 mol of a diatomic gas is taken reversibly around the cycle shown in the $T-S$ diagram of Fig. $20-35$, $\begin{array}{llll}\text { where } & S_{1}=6.00 & \mathrm{~J} / \mathrm{K} & \text { and }\end{array}$ $S_{2}=8.00 \mathrm{~J} / \mathrm{K} .$ 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 $(\mathrm{g})$ path $1 \rightarrow 2,(\mathrm{~h})$ 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?

Suppose 2.00 mol of a diatomic gas is taken reversibly around the cycle shown in the $T-S$ diagram of Fig. $20-35$, $\begin{array}{llll}\text { where } & S_{1}=6.00 & \mathrm{~J} / \mathrm{K} & \text { and }\end{array}$
$S_{2}=8.00 \mathrm{~J} / \mathrm{K} .$ 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 $(\mathrm{g})$ path $1 \rightarrow 2,(\mathrm{~h})$ 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

Ideal Gas Behavior and Degrees of Freedom

Ideal gas behavior assumes that the gas molecules do not interact except through elastic collisions and that their internal energy depends only on temperature. The number of active degrees of freedom, which can be affected by molecular structure such as translational and, under certain conditions, rotational or vibrational motions, determines the specific heat capacities. In cases where rotational and vibrational modes are inactive, the gas obeys simpler energy relations similar to a monatomic gas.

Thermodynamic Cycle Analysis

In a thermodynamic cycle, a system undergoes a series of processes and eventually returns to its initial state, meaning that the net change in state functions like internal energy over the cycle is zero. Analyzing such cycles involves carefully tracking the heat and work transfers in each process, utilizing the first law of thermodynamics to relate the energy exchanges within the cycle.

Adiabatic Process

An adiabatic process is characterized by the absence of heat exchange between the system and its surroundings. In such processes, any work done on or by the system directly results in a change in the system’s internal energy. This concept is key for understanding rapid processes or situations where insulation minimizes thermal interactions.

State and Path Functions

State functions, like internal energy and entropy, depend solely on the state of the system and not on how it arrived there, whereas path functions, such as heat and work, depend on the specific process taken. Recognizing the distinction between these concepts is crucial when evaluating energy changes in thermodynamic cycles.

Isothermal Process

An isothermal process is one that occurs at a constant temperature. For an ideal gas undergoing an isothermal change, the internal energy remains constant, meaning that the heat transferred into the system is exactly balanced by the work done by the system. This process has special analytical significance in thermodynamics, particularly when using equations that relate pressure, volume, and temperature.

First Law of Thermodynamics

The First Law of Thermodynamics expresses the conservation of energy in a system, stating that the change in internal energy is equal to the heat added to the system minus the work done by the system. This principle is essential in analyzing energy exchanges in cyclic and non-cyclic processes, linking heat transfer, work, and changes in a system’s internal state.

Temperature-Entropy Diagram

A Temperature-Entropy (T-S) diagram is a graphical representation of a thermodynamic process where temperature is plotted against entropy. It is useful for visualizing cycles, determining the heat exchanged in reversible processes (with the area under a process curve representing the heat transfer), and comparing different thermodynamic paths.

Transcript

00:01 In this problem on the topic of entropy, we are to suppose that two moles of a diatomic gas is taken reversibly around the cycle that is shown in the temperature versus entropy plot in the figure.

00:12 And we are given the value for s1 and s2 to be 6 joules per kelvin and 8 joules per kelvin respectively, where the molecules do not rotate and oscillate.

00:21 We want to find the energy that is transferred as heat q4, path 1 to 2, path 2 to 3, and the full cycle.

00:29 We then want to find the work w for the isothermal process, and if the volume v1 in state 1 is 0 .2 cubic meters, we want to find the volume in state 2 and then in state 3.

00:41 We then want to find the change in internal energy for path 1 to part 2, part 2 to part 2 to part 3, and for the full cycle.

00:49 Now, lastly, we want to find the work w for the adiabatic process.

01:00 So from equation 20 .1, we can infer that the heat transfer q is the integral of the temperature t times the infinitesimal entropy change ds.

01:13 Now this corresponds to the area under the curve in a ts diagram.

01:18 And so since the area of a rectangle is height times width, we have the heat energy or the heat transfer for path 1 to 2, which is q 1 to 2, to be 350 times 2 reading from the graph, which is a heat transfer of 700 joules.

01:48 Now for part b, we have no area under the curve for process 2 to 3, so we can conclude, therefore, that the heat transfer during this process, q 2 to 3 is equal to 0.

02:05 For part c, for the cycle, the net heat should be the area inside the figure.

02:10 So using the fact that the area of a triangle is a half base times height, we get the net heat transfer to be half times 2 times 50, which gives us a net heat transfer in the entire cycle to be 50 joules...

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