A sample of a monatomic ideal gas occupies 5.00 L at...

A sample of a monatomic ideal gas occupies $5.00 \mathrm{~L}$ at atmospheric pressure and $300 \mathrm{~K}$ ( point $A$ in Figure P21.63). It is heated at constant volume to $3.00$ atm (point $B$ ). Then, it is allowed to expand isothermally to $1.00 \mathrm{~atm}$ ( point $C$ ) and at last is compressed isobarically to its original state. (a) Find the number of moles in the sample. (b) Find the temperatures at points $B$ and $C$ and the volume at point $C$. (c) Assuming that the specific heat does not depend on temperature, so that $E_{\text {int }}=3 n R T / 2$, find the internal energy at points $A, B$, and $C .$ (d) Tabulate $P, V, T$, and $E_{i n t}$ at the states at points $A, B$, and $C .$ (e) Now consider the processes $A \rightarrow B, B \rightarrow C$, and $C \rightarrow A .$ Describe just how to carry out each process experimentally. (f) Find $Q, W$, and $\Delta E_{\text {int }}$ for each of the processes. (g) For the whole cycle $A \rightarrow B \rightarrow C \rightarrow A$, find $Q, W$, and $\Delta E_{\text {int }} .$

Key Concepts

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, is a fundamental relation in thermodynamics that connects pressure (P), volume (V), the number of moles (n), the universal gas constant (R), and temperature (T). It provides a simple model for the behavior of gases under various conditions and is crucial for solving problems involving changes in any of these state variables when the others are known.

Thermodynamic Processes

Thermodynamic processes describe how a system interacts with its surroundings when it undergoes changes in its state variables. Common processes include constant volume (isochoric), constant pressure (isobaric), and constant temperature (isothermal) transformations. Understanding these processes is critical for predicting how properties such as pressure, volume, and temperature will change under specific constraints.

Internal Energy of a Monatomic Ideal Gas

For a monatomic ideal gas, the internal energy is solely a function of temperature and is given by the equation E_int = (3/2)nRT. This concept is essential for calculating changes in the internal energy during various processes and connects the microscopic motion of particles to the macroscopic thermodynamic properties of the gas.

First Law of Thermodynamics

The first law of thermodynamics, which states that the change in internal energy (?E_int) is equal to the heat added (Q) to the system minus the work done (W) by the system (?E_int = Q - W), is fundamental for energy conservation in thermodynamic processes. It provides a framework for analyzing energy flow and transformations in processes such as heating, expansion, and compression.

Heat Transfer and Work in Processes

In any thermodynamic process, heat transfer (Q) and work (W) are path-dependent quantities that describe the energy exchanged between the system and its surroundings. Their values depend on the nature of the process (whether isothermal, isochoric, or isobaric), making it important to understand how to calculate these quantities for different types of processes in order to fully characterize the energy dynamics of a system.

Cyclic Thermodynamic Processes

Cyclic processes involve a system undergoing a series of transformations that eventually return it to its initial state. In such cycles, while the internal energy returns to its initial value (indicating no net change), the net heat and net work depend on the specific path taken. This concept is key when analyzing engines and refrigerators, where understanding the work done over a complete cycle is crucial.

Transcript

00:01 In this question, we are given a sample of monoatomic gas and it undergoes a cycle.

00:07 And it is described in this pv diagram, okay, starting with a, which is one atmospheres, 5 litres, okay, first step, first process it reaches on three atmospheres, and then undergo an isothermal expansion, until it goes to one atmosphere and then goes back to the original state.

00:36 Okay, so a, b, and c.

00:41 So there are seven parts in this question.

00:46 In part a, we want to calculate a number of modes of gas.

00:50 Okay, we are given that.

01:00 So at a, va is five liters.

01:07 P .a.

01:10 Is one atmosphere and then t .a.

01:16 Is 300 kelvin using the ideal gas equation.

01:28 P .b.

01:28 Equals to nrt.

01:36 Okay, so we can calculate the number of modes and it goes to p .v.

01:41 Divided by rt.

01:43 So one atmosphere is 1 .013 times 10 to the 5 pascal, 5 liters, 5 times 10 to the negative 3 meter cube.

01:51 Okay, r is 8 .31, t is 300 kelvin.

01:58 So the number of moles is 0 .203 modes.

02:03 So this is the answer for part a, and then part b, when to calculate the temperatures at b and c and volume at c okay so at b for process a the pressure goes from goes from one atmospheres to three amazheres so while volume is constant while v stays constant okay so we have from from pv equals to nrt we have pa over pb equals to ta over tb.

03:07 So tb is pb over pa times ta, we have 3 divided by 1 times 300, which is 900 kelvin.

03:20 Tc equals to tb, okay, goes to 900 kelvin because b2c is is isothermal process.

03:33 So these are the answers for temperature, tb and tc...

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