A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A in Fi

step 1 Hi friends, this is the problem based on kinetic theory of the gas. Here it is given 5 litre vol

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

A sample of monatomic ideal gas occupies $5.00 \mathrm{L}$ at atmospheric pressure and $300 \mathrm{K}$ (point $A$ in Figure $\mathrm{P} 21.67$ ). It is heated at constant volume to 3.00 atm (point $B$ ). Then it is allowed to expand isothermally to 1.00 atm (point $C$ ) and at last compressed isobarically to its original state. (a) Find the number of moles in the sample. (b) Find the temperature at points $B$ and $C$ and the volume at point $C$. (c) Assuming that the molar 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 $E_{\text {int }}$ for 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

Internal Energy of a Monatomic Ideal Gas

For a monatomic ideal gas, the internal energy is primarily a function of temperature and is given by the expression E_int = (3/2)nRT. This relationship emerges from the kinetic theory of gases, reflecting the translational degrees of freedom of the molecules and is crucial for connecting temperature changes to energy changes in the system.

First Law of Thermodynamics

The first law of thermodynamics is a statement of energy conservation, which asserts that the heat added to a system is equal to the increase in internal energy plus the work done by the system. This law provides the framework for analyzing energy transfers in thermodynamic processes and cycles by balancing heat, work, and changes in internal energy.

Isobaric Process

An isobaric process occurs at constant pressure. During such a process, the volume and temperature of the gas can change, and the work done can be calculated as the product of the constant pressure and the change in volume. This process is important in thermodynamics because it often approximates real-life scenarios where pressure remains nearly constant.

Thermodynamic Cycle

A thermodynamic cycle consists of a series of processes that eventually return a system to its initial state. In such cycles, while the internal energy change over a complete cycle is zero, net work and heat transfer can occur. Understanding thermodynamic cycles is essential for analyzing engines and refrigerators, as they highlight the interplay between different processes and the overall efficiency of energy conversion.

Constant-Volume Process

A constant-volume process, also known as an isochoric process, is one in which the volume of the gas remains fixed while other variables like pressure and temperature may change. In such a process, no work is done by the gas as work in thermodynamics is associated with volume change, and any heat added to the system directly alters the internal energy and temperature.

Ideal Gas Law

The ideal gas law relates pressure, volume, temperature, and the number of moles of a gas through the equation PV = nRT. It is a fundamental equation in thermodynamics that allows one to calculate any one variable if the others are known. This relationship is essential for understanding how gases behave under various conditions and is foundational to solving many problems involving gas processes.

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the process. For an ideal gas undergoing an isothermal transformation, the internal energy remains constant. The work done during an isothermal process is achieved by compensating changes in pressure and volume according to the ideal gas law, with heat exchange occurring to maintain a steady temperature.

Transcript

00:01 Hi friends, this is the problem based on kinetic theory of the gas.

00:04 Here it is given 5 litre volume of monotermic gas at initial pressure 1 atmosphere and initial temperature 300 kelvin considering this state to be a.

00:56 So volume at the state a is 5 litre at pressure.

01:03 1 atmosphere and temperature 300 kelvin.

01:17 A to b heated at constant volume.

01:27 So at b volume is 5 litre, pressure becomes 3 atmosphere.

01:52 V2c allowed to expand isothermally.

02:08 So at state c temperature at c is cut to temperature at b and pressure becomes one atmosphere and c2t compressed isomerically to its original state.

03:25 In the part a we have to calculate number of moles b part b have to find temperature temperature.

03:47 At b and c volume at b and volume at c in c part we have to find internal energy at points a b and c d tabulate pressure volume temperature internal energy for points a b and c discuss experiment to achieve process a to b to c and c to a a in effort we have to find amount of heat transfer work done and change in internal energy in each process g for complete cycle calculate or find q w and change in internal energy.

06:00 Let us start solving first part.

06:07 Number of moles can be defined as pb upon rt pressure one atmosphere volume is 5 liter gas constant 8 .31 4 temperature is 300 kelvin so it is to be 0 .203 more b part temperature at b will be temperature at a into pressure at b upon pressure at a because volume is constant 300 into 3 upon 1 that is 900 kelvin now temperature at c is called to temperature at b because the process is is isothermal it will be 900 kelvin volume at c will be equal to volume at a because pressure is constant.

07:44 Substitute the value 5 litre, 900 upon 300, so it will be 15 liter.

08:03 Internal energy at a 3x2, nr temperature at a, number of points, 0 .203, gas constant 8 .3, 4 temperature is 300 so it is to be 760 jule internal energy at point b 3 by 2 and our temperature at which will be also equal to temperature at c so internal energy at b is called to internal energy at c 3 by 2 .203, 8 .314.

09:15 So it is to be.

09:28 Now d part, we have to make a list state, pressure in atmosphere, volume in liter, temperature in kelvin and internal energy in kilojude.

10:20 A, v and c.

10:25 At a, pressure is one atmosphere.

10:27 Volume is 5 liter temperature is 300 kelvin energy is 0 .76 kilojoude at b pressure becomes 3 atmosphere volume remain 5 liter temperature becomes 900 kelvin energy will be 2 .28 kilojoude at c pressure 1 atmosphere volume becomes 15 liter temperature is 900 kelvin and energy will remain 2 .38.

11:08 E part.

11:16 For process a, crystal position is fixed for process bc, surrender to be kept and open at 900 kelvin for c2a, surrender, capped.

12:09 In a room at 300 kelvin and allowed to effort...

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