FIGURE:
h t t p : / / g o o . g l / a 7 j C F
The figure is a PV diagram for a reversible heat engine in which 1.0 mol of argon, a nearly ideal monatomic gas, is initially at STP (point a). Points b and c are on an isotherm at T = 423 K. Process ab is at constant volume, process ac at constant pressure.
What is the efficiency of this engine?
The efficicy is (Integral PdV)/(Interval TdS) for the cycle.
To find the efficiency of the heat engine, we need to use the given information and the formula for efficiency of a heat engine:
Efficiency = 1 - (Tc / Th)
where Tc is the temperature at the end of the heat sink (in this case, point c) and Th is the temperature at the end of the heat source (in this case, point b).
In the provided figure, we can see that the points on the diagram represent different states of the gas. The process ab is at constant volume, meaning the volume of the gas remains constant during that process. On the other hand, process ac is at constant pressure, meaning the pressure of the gas remains constant during that process.
Since point b and c are on the same isotherm (constant temperature line), we can assume that the temperature is the same for both states, which is T = 423 K.
We need to find the temperatures at point b and c on the PV diagram. However, the given link is not accessible, so I am unable to see the diagram.
To find the temperatures at points b and c, you need to look at the PV diagram and find the corresponding values of pressure (P) and volume (V) at those points. Once you have the pressure and volume values, you can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation, we can solve for T:
T = PV / (nR)
By substituting the values of pressure and volume at points b and c into the equation, you can calculate the temperatures at those points. Once you have the temperatures, you can calculate the efficiency using the formula mentioned earlier.