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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