One
mole of a monatomic ideal gas is taken from an initial state of
pressure
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Solution: Click For PDF Version
We are
considering one mole of a monatomic ideal gas as it undergoes
thermodynamic processes as shown in the diagram. Its initial state
is at the point a in
the diagram. At a
pressure of the gas is
(1)
As the gas
undergoes isothermal expansion at temperature
As the
process is isothermal there is no change in the internal energy of
the gas, therefore, the heat absorbed by the gas,
The change in entropy
(2) The process b to d takes place at constant volume. Therefore, no work is done on the gas during this process. Therefore, the heat absorbed by the gas during 2 will be
We have used that for a monatomic gas
and at b
And at d
And
The change in entropy of the gas during the process 2 will be
The change in entropy from a to d in the combined processes 1 and 2 will be
(3)
We consider
part 3 process. It is an isothermal compression at temperature
and
Therefore,
(4)
In the
process 4 the gas is subjected to an isobaric expansion from volume
Therefore, the work done on the gas during the path 4 will be
And the heat absorbed by the gas during 4 will be
We have used that for an ideal monatomic gas
And
The change in entropy
Therefore,
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and volume
to a final state of pressure
by two different processes. (1) It
expands isothermally until its volume is doubled, and then its
pressure is increased at constant volume to the final state.
(2) It is compressed isothermally until its
pressure is doubled, and then its volume is increased at constant
pressure to the final state. We have to show the path of each process
on a pV diagram. For each process we have to calculate in terms of
;
and (d) the change in
entropy of the gas,
from
volume
from a to b
along the path 1, the work done on the gas will be
, during the process 1 will be
.
Therefore, the work done on the gas
