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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