An
engine carries 1.00 mol of
an ideal monatomic gas around the cycle shown in the figure. Process
AB takes place at
constant volume, process BC is adiabatic, and process CA takes place
at constant pressure. (a) We
have to compute the heat Q, the change in internal energy
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Solution: Click For PDF Version Data of the problem are
Amount of
ideal monatomic gas is
We use the ideal gas equation for determining the values of the thermodynamic state variables at points A, B, and C.
And
(a) Process AB During the process AB the work done on the gas will be zero, as the change occurs at constant volume. Change in the internal energy of the gas during AB is determined by the change in temperature, that is
From the first law we get
Process BC
As the change
from B to C is adiabatic, the work done on the gas,
As the gas is
monatomic
Therefore,
Process CA The process CA takes place at constant pressure. The work done on the gas will be
From the first law we calculate the heat absorbed by the gas during this process. We find
We now have the data for calculating the work done and the heat absorbed during the cycle as a whole.
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,
and the work W for each of the three processes and for the cycle as a
whole. (b) If the
initial pressure at point A is 1.00 atm, we
have to find the pressure and the volume at points B and C. We will
use
and
and
,
can be obtained from the relation
And
as the process is adiabatic
.
Change
in internal energy of the gas will be
Total
heat absorbed by the gas during the whole cycle will be