Showing posts with label Isothermal Process. Show all posts
Showing posts with label Isothermal Process. Show all posts

Thermodynamics Problems with Solutions Five

We are solving series of problems based on the concepts of thermodynamics. We also deal about calorimetry in this chapter which deals about conversion of heat energy to other forms of energies and its applications further. We do define terms like specific heat and latent heat to explain this properties and they are the basic terms of heat concepts. When there is a change in the temperature, we need to deal with specific heat concept and when there is a change of state, we need to study it in terms of latent heat and during this process, all the supplied heat energy is used to change the state of the system and hence its temperature remains constant.

Problem

During an adiabatic process,pressure of the gas is proportional to the cube of the temperature and basing on that we need to find the ratio of specific heats of the gas. Problem is as shown in the diagram below.


Solution

We need to take the relation between pressure and temperature and taking that into consideration with the given data, we can get the relation between pressure and temperature in the adiabatic process and the problem can be solved as shown in the diagram below.


Problem

A metal sphere of known radius and specific heat is given to us and it is rotating about its own axis with certain rotations per second. When it is stopped half of its energy is converted into heat and we need to measure the raise in the temperature of the system and the problem is as shown in the diagram below.


Solution

As the body is rotating it has rotational kinetic energy and half of it is converted into heat energy as per the given problem. Taking law of conservation of the energy, we can solve the problem as shown in the diagram below.



Problem

The relation between internal energy,pressure and volume is given to us as shown in the diagram below. We need to find the ratio of specific heats and some constants are also available in the problem.


Solution

We need to differentiate the given equation to get the change in internal energy and hence it can be expressed in terms of specific heat of the gas at constant volume. Problem can be further solved as shown in the diagram below.


Problem

Work done by a system under isothermal conditions has to be determined that change its volume from one to other and satisfy the given equation. Problem is as shown in the diagram below.


Solution

Relation of pressure with other physical quantities is given to us as shown in the diagram below. We need to measure the work done and the pressure is not constant here. So to get the work done we shall integrate the pressure with the change in the volume as shown in the diagram below. By simplifying the equation further, we can get the solution as shown in the diagram below.


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Thermodynamics Problems with Solutions Four

We are solving series of problems based on the concept of thermodynamics. It is a branch of physics that deals with the heat energy conversion into other forms and its applications. Carnot engine is a ideal heat engine that converts the given heat into work under a cyclic process. As the process is cyclic process, internal energy remains same and all the supplied heat is going to be converted into work. But we know that heat is a disordered format of energy and it cannot be completely converted into work. That is the reason why, heat engine cannot have hundred percent efficiency. Heat engine will have three basic parts by name source, working substance and sink. Source do supply heat, working substance do convert the heat into work and the extra generated heat to the sink.

Problem

Pressure and volume graph for two different gases during adiabatic process is given to us as shown in the diagram below. We need to know which graph belongs to which gas.


Solution

From the pressure and volume graph, it can be found that the slope of the graph is directly proportional to the ratio of specific heats of the gas. It is clear from the graph that the slope of the second curve is more and hence it shall be the gas with more ratio of specific heat value. Solution of the problem is as shown in the diagram below.


Problem

A refrigerator is placed in a room of temperature of 300 kelvin and the system temperature is 264 kelvin. We need to measure the how many calories of heat shall be delivered to the room to the room for each kilo kelvin of energy consumed by refrigerator by the system ideally. Problem is as shown in the diagram below.


Solution

We know that proficiency of a refrigerator is the amount of work done when compared with the heat energy supplied to the system. It can also be expressed in terms of temperature as shown in the diagram and is solved as shown in the diagram below.


Problem

In carnot’s engine efficiency is 40 % for a certain temperature of the hot source. To increase the efficiency of the system by 50 %, what shall be the source temperature and the problem is as shown in the diagram below.


Solution

We know that the efficiency of heat engine is the magnitude of the work done when compared with the heat energy supplied and it can be expressed in terms of absolute temperature as shown in the diagram below.

We need to apply it for two cases and solve the problem as shown in the diagram below.


Problem

The relation between temperature and pressure is given to us for a certain gas and we need to find the specific heats ratio of that gas. Problem is as shown in the diagram below.


Solution

This process is adiabatic process and we need to write the given relation between pressure and temperature from the given format to the standard format so that we can solve and find the ratio of specific heats of the given gas as shown in the diagram below.


Problem

When a mono atomic gas expands at constant pressure, the percentage of the heat supplied that increases the temperature of the gas and in doing external work is how much is the problem and it is as shown in the diagram below.


Solution

We know that if volume is kept constant, work done is zero and corresponding specific heat is defined as per it and vice versa. When we write ratio of specific heats, it becomes the ratio of heat supplied to the change in the internal energy and internal energy change the temperature of the system. Thus we can find the percentage value basing on the specific heats ratio value as shown in the diagram below.



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Thermodynamics Problems with Solutions Three

We are solving series of problem in the topic thermodynamics. There are different types of process in thermodynamics. Isobaric process means pressure of the system is kept constant. Isothermal process means temperature of the system is kept constant. Adiabatic process means heat energy of the system is kept constant. In each case, work done is different and we need to measure as per the given process. Zeroth law of thermodynamics is regarding temperature concept. First law of thermodynamics is regarding conservation of energy and second law of thermodynamics is regarding direction of flow of heat.

Problem

Three samples of same gas has initially same volume. Volume of each one is doubled and the process is different in each case. It is adiabatic, isobaric and isothermal respectively . if all the final pressures are equal, we need to know the initial pressures of the three and the problem is as shown in the diagram below.


Solution

The relation between pressure and volume is different in each case. Isobaric process means pressure remains constant and in that case, initial pressure is equal to final pressure. In the case of isothermal process, as temperature is constant, we can apply boyle’s law and find the final pressure in terms of initial pressure. Adiabatic process get the ratio of specific heats into the picture and taking that into consideration, we can solve the problem as shown in the diagram below.


Problem

The pressure inside a tyre at one temperature is given to us as four atmospheric pressure and we need to find the temperature when the tyre bursts suddenly. Problem is as shown in the diagram below.


Solution

As the tyre bursts suddenly, it is going to be a adiabatic process and the heat energy of the system remains constant. We need to apply and find the relation between temperature and pressure using the pressure and volume relation of the adiabatic process and simplify the problem as shown in the diagram below.


Problem

There are two cylinders with the same ideal gas at the same temperature. There are pistons on both of them and their initial temperature is same. If one piston is allowed to move freely and the other is fixed, we need to compare the temperature of the second case when compared with the first case. Problem is as shown in the diagram below.


Solution

In the first case pressure is kept constant and in the second case it is the volume that is kept constant by sealing the system. Taking the respective specific heats and using the equation for the heat energy in each case, we can solve the problem as shown in the diagram below.


Problem

An ideal gas after going through four thermodynamic states has come back to its initial state. Heat energy in each case is given to us and work done in three cases is given to us as shown in the diagram below. We need to find the work done in the fourth case.


Solution

As the process is cyclic, there is no change in the internal energy of the system and as per the first law of thermodynamics, the heat energy supplied in this case will be the work done itself. By equating the total heat energy with proper sign to the total work done, we can solve the problem as shown in the diagram below.


Problem

PV diagram of an ideal gas is as shown and we need to measure the work done during a part as shown in the diagram below.


Solution

Work is said to be done when there is a change in the volume of the system other wise the work done is zero. In a PV diagram, work done is the area under the graph. We can use the shape and find the area of the graph to solve the problem.



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Thermodynamics Problems with Solutions One

Thermodynamics Problems with Solutions Two

We are solving series of problems on the topic thermodynamics. We need to be familiar with the concepts like specific heat and latent heat. Specific heat is the amount of heat energy required to raise the temperature of unit mass of substance by one degree centigrade or kelvin. In the case of gases, we get two different kinds of specific heats one at constant pressure and the other at constant volume. Latent heat is the amount of heat energy required to convert unit mass of substance from one state to other and it happens at constant temperature. As the heat energy is also conserved, the heat lost by a hot body is the heat gained by the cold body when the bodies are connected and there is no loss of heat energy in any format.

Problem

The ratio of specific heats of has is given to us in the problem. We need to measure the change in the internal energy of the system when volume gets doubled at constant pressure. Problem is as shown in the diagram below.


Solution

We can write the value of internal energy in terms of first law of thermodynamics and work done can be written as  the product of pressure and change in volume of the system. It can be further simplified as shown in the diagram below.


Problem

One mole of mono atomic gas is mixed with one mole of diatomic gas are mixed. We need to find the effective specific heat of the system and the problem is as shown in the diagram below.


Solution

We can write the equation for the effective specific heat by taking the conservation of energy into consideration as shown in the diagram below. We need to apply the value of specific heat of mono and diatomic gas values as per the standards and solve the problem as shown below.


Problem

Certain amount of heat is required to raise the temperature to a certain value at constant volume and we need to find out the heat energy required to raise the temperature of the system at constant pressure and the problem is as shown in the diagram below.


Solution

When the volume is constant, to find the heat energy required to raise the temperature to a certain value, we need to use the definition of specific heat of gas at constant volume. To find the heat required to raise the temperature of the system at constant pressure, we need to use the definition of specific heat of gas at constant pressure. The detailed solution is as given in the diagram below.


Problem

One mole of ideal gas expands to double to its volume under isothermal conditions at a given constant temperature as shown in the diagram below. We need to measure the work done in this case.


Solution

We know that isothermal process means the work is done at constant temperature. Here pressure is not constant and to measure the work done, we shall integrate the pressure impact under the volume conditions and we can derive a mathematical equation as we have learned. Taking that into consideration, we can solve the problem as shown in the diagram below.


Problem

For an adiabatic expansion, if a monoatomic gas expands by a certain percentage, we need to measure the percentage variation in its pressure and the problem is as shown in the diagram below.


Solution

We know that in adiabatic process, the heat energy of the system is constant and it is not going to change. Here temperature is not constant and hence Boyle’s law is not valid here. The relation between pressure and volume are different and taking that into consideration, we can solve the problem as shown in the diagram below.




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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