Showing posts with label Expansion of Gases. Show all posts
Showing posts with label Expansion of Gases. Show all posts

Expansion of Gases Problems with Solutions Three

We are solving series of problem on the concept of expansion of gases. Pressure of the gas varies proportionality with the variation of temperature. It is also true that the variation of volume is proportionate to the variation of temperature at constant volume. Basing on this concepts,we can define different scales of temperatures and different gas thermometers. We also have ideal gas equation according to which the product of pressure and volume is directly proportional to the absolute temperature. The constant there is the product of number of moles and universal gas constant. It is constant for all the kinds of gases all over the universe.

Problem

Variation of volume of the gas with absolute temperature is given to us as shown in the diagram below. We need to find the what happens to the pressure during this process.


Solution

As the graph is a straight line passing through the origin, pressure is constant and hence it is not a variable at any point. The solution is as shown in the diagram below.


Problem

Variation of pressure with volume is shown in the graph for two different temperatures. We need to find the relation between the temperateness.


Solution

The variation of pressure with volume is as shown in the diagram below. We know that the product of pressure and volume is equal to the product of universal gas constant and absolute temperature. Thus the slope for the second temperature is more and hence that temperature is more.


Problem

During an experiment an ideal gas is following a certain rule as given in the problem diagram. The initial temperature and volume of the gas is given to us. If volume of the gas is doubled, we need to know what happens to the temperature.


Solution

As the gas is ideal gas equation, it satisfy the ideal gas equation also. We can use the pressure value from ideal gas equation and substitute it in the given relation of the problem. Then it can be proved that the volume and temperature are inversely proportional to each other. Then the problem can be solved as shown in the diagram below.



Problem

A horizontal tube of one meter length has uniform area of cross section and it has a mercury pellet of length 10 centimeter at the middle. The temperature and pressure at one end are given as shown in the diagram below. If we know the pressure at the other end and the problem is as shown in the diagram below.


Solution

We need to apply ideal gas equation to solve the problem. As area of cross section of the system is constant, we can write the volume as the product of area of cross section and length of the air column and volume is proportional to the length of the air column. As temperature at one end increases air column there expands and pushes the mercury pellet to the other side. Thus on the higher temperature side there will be more air column and vice versa. By applying the ideal gas equation, we can solve the problem as shown in the diagram below.


It is further simplified as shown in the diagram below.



Problem

We need to find out at what temperature RMS velocity of hydrogen gas molecule is equal to the RMS velocity of the oxygen gas molecule at the given temperature and the problem is as shown in the diagram below.


Solution

We know that RMS velocity depends on the nature of the gas but in this case both hydrogen and oxygen gas are diatomic in nature and it is not a factor here. It is inversely proportional to the molecular weight and it is 2 for hydrogen and 32 for oxygen. We can write the formula for the RMS velocity and solve the problem as shown in the diagram below.



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Expansion of Gases Problems with Solutions Two

We are solving series of problems based on the concept expansion of gases. When heat energy is given expansion of gas is much bigger in scale than that of solids and liquids due to weak molecular force of attraction among the gas molecules. Expansion of gases has to be studied taking the impact of pressure on it also as it plays a significant role. We have different gas laws to study their motion. A gas is said to be ideal when it satisfy all gas laws at all temperatures and pressures. But practically no gas is ideal and the existing gases are called real gases. They obey gas laws only at high temperatures and low pressures.

Problem

Density of a gas at known temperature and pressure is given to us as shown in the diagram below. We need to find the density of the system at different pressure and temperature.


Solution

We can rewrite the ideal gas equation in terms of density. Here we write volume of ideal gas as the ratio of mass and density and mass of the gas is constant. We can solve the problem as shown in the diagram below.


Problem

It is given in the problem that at two different temperatures gas occupies different volumes and we need to find the volume expansion coefficient of the gas and the problem is as shown in the diagram below.


Solution

We know that at constant pressure, volume of the gas is directly proportional to the initial volume and change in the temperatures. Basing on that we can define coefficient of volume expansion of the gas as the ratio of change in the volume of the gas with its original volume per unit change in the temperature. Basing on that we can solve the problem as shown in the diagram below.


Problem

It is given in the problem that in a constant volume gas thermometer, gas is under different pressures at two different temperatures and we need to find the pressure of a gas at a different temperature. Problem is as shown in the diagram below.

 

Solution

We know that gas expansion cannot be studied under three parameters volume,temperature and pressure. Hence we keep one parameter constant and study the other two. Here volume is kept constant and it is found that change in the pressure is directly proportional to the initial pressure and change in the temperature. Basing on that coefficient of pressure expansion of gases is defined. Given data is substituted in the formula and is solved as shown in the diagram below.


Problem

The length of air column and mercury thread in a quill tube is given to us and it is given in the problem that the open end is in the upward direction. When the tube is tilted by a certain angle with the horizontal,we need to find the length of the air column. Problem is as shown below.


Solution

We know that as the temperature of the system is constant, we can apply Boyle’s law to solve the problem. Here in the place of volume we can write the product of area of cross section of the tube and length of the pipe. Area of cross section is constant here and hence the law can be rewritten as the product of length and pressure is constant for a given gas. Basing on that the problem is solved as shown in the diagram below.



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Expansion of Gases Problems with Solutions One

We are solving series of problems based on the concept expansion of gases. When heat energy is given to the gas, they get the heat energy and it spreads the gas molecules quire far away and they fail to come back to their original position and it leads to expansion. Gas is not having linear and areal expansion and there is no particular shape and size. While we are studying expansion of solids and liquids, we have not worried about the impact of pressure as its impact is negligible and hence we had not taken that into consideration. But in the case of gases, impact of pressure is big and it can not be ignored. 

There are three parameters here with gases like pressure,volume and temperature and we can not study all three simultaneously. Hence we keep any one parameter constant and study the other two parameters. If pressure is kept constant, volume varies proportionate with temperature. If volume is kept constant, pressure of the gas is directly proportional to its absolute temperature and basing on that we can define coefficients of expansions of the gas. It is proved that both the coefficient of expansions are same and it is the same who is kept constant and who is variable is not going to make any difference.

Problem

What shall be the percentage change in the pressure of the gas when percentage change in the volume of the gas is five percent at constant temperature and the problem is as shown in the diagram below.


Solution

As temperature is constant for a given gas, we can apply Boyle’s law. According to this rule, product of volume and pressure of a given mass of the gas is constant at the constant temperature. By applying the data of the problem as shown in the diagram below and solve it.


Problem

An air bubble doubles its radius on reaching the top of a water lake at constant temperature and we need to know the depth of the water. Problem is as shown in the diagram below.


Solution

Here also we are going to apply Boyle’s law. We shall assume that the temperature of the system is constant in this case. When the bubble is at the bottom of the lake there is pressure on it due to water as well as atmosphere. When the bubble reaches the top there is no water on its top and hence there is pressure only due to atmosphere. Hence pressure on the top is less and hence the volume will be more. By applying Boyle's law, we can solve the problem as shown in the diagram below.


Problem

For an ideal gas volume and temperature at two different constant pressures as shown in the diagram below. We need to find the relation between two pressures.


Solution

At constant pressure we can get the relation between volume and pressure basing on the Charles law. The problem can be solved as shown in the diagram below.


Problem

Two identical containers are connected by a capillary tube contain air at NTP conditions. If one of the container is immersed in hot boiling water, what is the new pressure. Problem is as shown in the diagram below.


Solution

As the volume of the system is constant, we need to apply the relation that the pressure is directly proportional to the absolute temperature. Initially the two have different pressures. Temperature of one system is changed in the problem and they together get a common pressure. Problem is solved as shown in the diagram below.


Problem

Two spheres of different volumes are connected with a capillary tube of negligible volume as shown in the problem below. They contain ideal gas at given temperature and pressure conditions. Keeping temperature of one constant and if the temperature of the second is increased, we need to find the final pressure.


Solution

As the number of moles of the gas is constant in the given case, we can apply ideal gas condition to solve the problem. When temperature of one system is changed , the total pressure of the system gets effected and we need to apply ideal gas condition as shown in the diagram below to solve the problem.



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Thermal Properties of Heat Complete Lesson

Heat is a disordered form of energy and to know the direction of flow of the heat, we need the concept of temperature. Temperature is a measure of the heat energy. Heat is measured with a unit called calorie and temperature is measured with kelvin in the standard international system. The rise in the temperature of a body causes the expansion in general in the body and it is called thermal expansion. To study this expansions, we need coefficients of expansions and they were separately defined for solid state, liquid state and gaseous state materials. 

Heat flows from one place to other in different ways called conduction, convection and radiation. Conduction need a solid medium, convection need a fluid media and radiation is not in need of any media for the propagation.

Here in this lesson we have discussed about all this topics in detail.

Expansion of Solids and Applications
                                                     

Anomalous expansion of water 

 Expansion of Liquids Problems with Solutions

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Thermodynamics and First Law of Thermodynamics

Thermodynamics deals with the heat energy and its transformation to other formats of energy’s.It deals with different formats of heat energy is, conversion and its applications.

Coordinates of Thermodynamics of the fundamental physical quantities basing on which thermodynamics is steadied.Temperature, pressure, volume, internal energy and entropy are some of the coordinates of Thermodynamics.

Temperature is a measure of heat energy. Body with high heat energy will have more temperature than the body with a low heat energy.

Pressure is defined as the force acting per unit area.

Internal energy is the energy that the system poses inside. In the case of a gas internal energy is the sum of potential energy due to the position of the gas molecules, translational kinetic energy you to the motion of the gas molecules, vibrational kinetic energy due to the oscillation of particles and rotational kinetic energy to do the rotation of molecules.

Depending on the nature of the gas either all of these energies or some of the energies are going to contribute for the internal energy.

We cannot measure the absolute energy of a body or a system and we can only measure the change in it.

Internal energy cannot do any external work.



There are some fundamental laws of thermodynamics. Among them  zero'th  law thermodynamics is the first one.

Definition

If two bodies A and B are in thermal equilibrium with a third body C separately, then  A and B are also in thermal equilibrium.

Work done by a gas

Consider a cylinder having a fixed area of cross-section. Let a gas at a pressure P is available in the system where the three sides of the system are closed and the fourth side is a movable piston.

Because of the expansion of the gas heater place a certain force on the piston and most the piston in the forward direction by a small distance. Basing on the deformation of the work done we can calculate the value as shown below.



If the volume increases in this process then the work done is treated as positive and vice versa.

Work done by the system is treated as positive and the work done on the system is treated as negative. Work done by the system produces in the expansion in the system and vice versa.

If the pressure is variable we shall integrate the equation to get the work done.

First law thermodynamics

The heat energy is given to a system is equal to the sum of change in its internal energy and the external work done by it.

This is nothing but law of conservation of energy. Energy is neither created, nor destroyed and it just converts from one format to another format. Heat the heat energy supplied is converted into the format of increase in internal energy and the external work done.

While applying this la we shall apply certain sign convention.

Heat given to the system is treated as positive and heat given by the system is treated as negative.

Increase in internal energy is treated as positive and a decrease in internal energy is treated as negative.

Work done by the system is treated as positive and the work done on the system is treated as negative.

We are going to solve small problem basing on the concept of first law thermodynamics where we need to apply proper sign convention.


Problem and solution

In a thermodynamic process, the pressure of a fixed mass of a gas is changed. In this process gas releases 20 is also of heat energy and trade jolts of work is done on the gas. If initial internal energy of the system is 30 J  what is the final internal energy?

We can solve this problem basing on the first law thermodynamics. According to the law the total heat energy supplied is equal to the sum of change in internal energy and the external work done.

As the system releases the heat energy shall be treated as negative.

Change in internal energy has to be calculated.

As the work is done on the system it shall be treated as negative.


The above diagram consists of another problem also. That problem is also solved basing on the same concept.

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Wave Motion an introduction 
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