## Thursday, January 12, 2017

### 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.