Figure 20-27 shows a reversible cycle through which 1.00 mol of a monatomic ideal gas is taken.

Figure 20-27 shows a reversible cycle through which 1.00...

Figure $20-27$ shows a reversible cycle through which $1.00 \mathrm{~mol}$ of a monatomic ideal gas is taken. Assume that $p=2 p_{0}, V=2 V_{0}, p_{0}=$ $1.01 \times 10^{5} \mathrm{~Pa}$, and $V_{0}=0.0225 \mathrm{~m}^{3} .$ Calculate (a) the work done during the cycle, (b) the energy added as heat during stroke $a b c$, and $(\mathrm{c})$ the efficiency of the cycle. (d) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures that occur in the cycle? (e) Is this greater than or less than the efficiency calculated in (c)?

Key Concepts

Efficiency of Heat Engines

The efficiency of a heat engine is defined as the ratio of the net work output to the heat input during the cycle. This metric quantifies the performance of the engine and indicates the fraction of the input energy that is converted into useful work, serving as a key parameter in comparing different thermodynamic cycles.

Heat Transfer and the First Law of Thermodynamics

The first law of thermodynamics states that energy is conserved in thermodynamic processes, meaning that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In cyclic processes, calculating the heat transfer in various strokes is essential to determine the overall energy balance and efficiency.

Carnot Efficiency

Carnot efficiency represents the maximum possible efficiency that a heat engine operating between two thermal reservoirs can achieve. It is calculated based on the highest and lowest temperatures of the cycle and serves as an ideal standard, setting an upper limit on the efficiency of real-world engines.

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure, volume, and temperature of an ideal gas. It is fundamental in describing and predicting the behavior of monatomic ideal gases in various thermodynamic processes and cycles, enabling the calculation of state variables during expansion and compression.

Reversible Cycle

A reversible cycle is a thermodynamic process that can be reversed by an infinitesimal change in a variable without leaving any net change in both the system and the surroundings. In such cycles, all processes occur infinitely slowly, ensuring the system is always in equilibrium, which allows for maximum efficiency and serves as an ideal benchmark in thermodynamics.

Work in Thermodynamic Processes

Work in thermodynamics, particularly in a pressure-volume (P-V) context, is calculated as the area under the curve in a P-V diagram during a process. In cyclic processes, the net work done over one complete cycle is given by the closed loop area, and understanding this concept is crucial for evaluating the performance of heat engines.

Transcript

00:01 Now here in this problem we are given a reversible cycle which is represented by this rectangle on pv diagram and the coordinates at point a we have v0 p -0 and the direction of this cycle is like this in the clockwise direction and the coordinates at point c, this point is c is v comma p.

00:24 Now there are five parts in the equation.

00:27 We'll be solving the a part first and in the a part we have to calculate the one.

00:33 Work done during the cycle now this cycle is represented on the pv diagram so if you calculate the area of this rectangle the area of this rectangle would represent the work done this is the concept that we studied in thermodynamics so the work done of the rectangle will be equals to so let this be length and let this be height so length multiplied with height so length we have v minus v0 this is the length minus sorry multiplied with height so height we have p minus p not now in the equation it is given that p is equal to two times p not and v is equals to two times v0 so we can plug it here so that would be two v0 minus v0 multiplied with two p knot minus p not so that will be equals to v0 times p not the value of v0 in the equation is given to be as 0 point 2 0 .025 multiplied with p .0 which we have 1 .01 times 10 to the power 5 pascal so from here the work done comes out to be as nearly 2 .27 times 10 to the power 3 jule so this is the answer for the a part now next we're going to solve the b part of the problem where we have to calculate the energy added as heat during stock a, b, c.

02:23 So during the process a to b and then b to c, we have to calculate the heat added.

02:28 So the heat added will be q equals to heat added during the process a to b plus the heat added during the process b to c.

02:39 Now the process a to b is a constant volume process.

02:44 So for constant volume process we have q equal...

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