One
mole of an ideal diatomic gas is caused to pass through the cycle
shown on the pV diagram, where
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Solution: Click For PDF Version We are considering thermodynamic cyclic processes of one mole of a diatomic ideal gas. For a diatomic gas the ratio of specific heats .
Three states of the gas marked by labels 1,2, and 3 are connected by thermodynamic processes shown in the pV diagram. Pressure, volume, and temperature of the gas at the point 1 are
and
Using the ideal gas equation of state, we note that the values of volume, pressure, and temperature at 2 are
and
Volume at 3 is
States 1 and 3 are connected by an adiabatic process. Therefore,
And,
We have
already noted that for a diatomic ideal gas
Process connecting the states 1 and 3 is isothermal. Therefore,
Work done on the gas will be
And heat absorbed by the gas during this part of the cycle will be
Change in entropy of the gas between the states 2 and 1 will be
Process
joining states 3 and 2 takes place at constant volume. Therefore, no
work is done on the gas during this process. Heat absorbed by the
gas,
Work done on the gas during the adiabatic process will be
And,
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We have to determine in terms of
and R: (a)
and
,
and (b) W, Q,
,
and
for all the three processes.
,
,
is equal to the change in internal energy between these two states.
Change
in entropy
States
1 and 3 are connected by an adiabatic process. In an adiabatic
process there is no exchange of heat energy. That is
Therefore,
