The study of thermodynamic cycles is like reading the heartbeat of an engine. Every curve on a P-V diagram tells a story of energy, work, and heat. In this problem, we are presented with a fascinating three-step cycle of a monoatomic ideal gas. Let's embark on this journey and decode the physics step by step.
Analyzing the Setup
We are given a quasi-static cycle consisting of three distinct processes: an isothermal expansion (a→b), an isochoric pressure drop (b→c), and an adiabatic compression (c→a).
Because the gas is monoatomic, we immediately know its specific heat capacities and adiabatic index. The molar heat capacity at constant volume is Cv=23R, and the adiabatic index is γ=35. These constants will be our primary tools.
Process a→b
The Isothermal Expansion
During the first leg of our journey, the gas expands from volume V1 to V2 while maintaining a constant temperature Ta=Tb.
For an ideal gas, if the temperature is constant, the internal energy remains constant (ΔU=0). According to the First Law of Thermodynamics, any work done by the gas must be perfectly balanced by heat absorbed from the surroundings.
The heat input during this isothermal process is given by:
Qin=Qab=nRTaln(V1V2)
Process b→c
The Isochoric Drop
Next, the gas undergoes an isochoric process. The volume is locked at V2, meaning the gas does zero work (W=0).
As the pressure drops from Pb to Pc, the temperature must also drop from Tb (which is Ta) to Tc. A drop in temperature means a decrease in internal energy, which is expelled as heat to the surroundings.
The heat released is:
Qout=−Qbc=−nCv(Tc−Ta)=nCv(Ta−Tc)
Process c→a
The Adiabatic Compression
Finally, the gas is compressed back to its original state at volume V1 without any heat exchange (Qca=0).
This is an adiabatic process, governed by the beautiful relation TVγ−1=constant. Applying this to points c and a, we get:
Since γ=35, the exponent γ−1 becomes 32. This gives us a direct bridge between the temperatures and volumes:
Evaluating the Options
Now, let's put our derived equations to the test against the given options.
Option A: We are asked to compare the heat released (Qout) to the heat absorbed (Qin). Notice that the cycle is drawn clockwise on the P-V diagram. A clockwise cycle means the net work done by the gas is strictly positive (W>0).
By the First Law of Thermodynamics for a complete cycle, the net heat exchange equals the net work done:
This immediately implies that Qin>Qout. The heat released is indeed smaller than the heat absorbed, regardless of the volume ratio. Option A is correct.
Option B: Let's construct the efficiency equation. Efficiency η is defined as the ratio of net work to heat input:
Substituting our expressions for Qout and Qin:
η=1−nRTaln(V2/V1)nCv(Ta−Tc)=1−RCvln(V2/V1)1−Tc/Ta
Notice how the initial temperature Ta factors out completely! Since Tc/Ta depends only on the volume ratio V2/V1, the entire efficiency expression is independent of the isothermal temperature Ta. Option B is correct.
Option C: We are given a specific volume ratio: V2/V1=8. Let's plug this into our adiabatic temperature relation:
Since 8=23, we have (23)2/3=22=4.
Therefore, Ta=4Tc. The temperature at a is exactly 4 times the temperature at c. Option C is correct.
Option D: Finally, let's check the pressures at a and b. Since process a→b is isothermal, Boyle's Law applies:
Rearranging for the pressure ratio:
The pressure at a is 8 times the pressure at b, not 4 times. Option D is incorrect.
The Final Verdict
Through a rigorous application of the First Law of Thermodynamics and the specific gas laws for isothermal, isochoric, and adiabatic processes, we have successfully navigated the cycle. The correct statements are A, B, and C.
