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 , 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
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Solution: Click For PDF Version Data of the problem are and 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, , can be obtained from the relation
As the gas is monatomic Therefore, And as the process is adiabatic .
Process CA The process CA takes place at constant pressure. The work done on the gas will be Change in internal energy of 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. Total heat absorbed by the gas during the whole cycle will be
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