Thermodynamic cyclic process - 2 isobaric and 2 adiabatic

In summary, a thermodynamic cyclic process consisting of two isobaric (constant pressure) and two adiabatic (no heat transfer) processes involves a working substance that undergoes a series of transformations. The isobaric processes allow for heat exchange with the surroundings while maintaining constant pressure, leading to volume changes. The adiabatic processes, on the other hand, involve changes in temperature and pressure without heat exchange, relying solely on work done on or by the system. Together, these processes create a cyclic path that enables the conversion of heat into work in heat engines, illustrating fundamental principles of thermodynamics.
  • #1
Thermofox
144
26
Homework Statement
A heat engine that uses an ideal gas carries out a cycle composed by:
##A\rightarrow B## isobaric expansion
##B\rightarrow C## adiabatic expansion
##C\rightarrow D## isobaric compression
##D\rightarrow A## adiabatic compression
All processes are reversible. I know that ##P_{AB}= 20 atm## ; ##P_{CD}= 10 atm## and that ##V_A= 5 L## ; ##V_B= 10 L##.

Assuming that the heat engine is able to work with a mole of a monoatomic and a mole of a diatomic gas determine:
1) The total heat absorbed by the gasses in a cycle
2) The minimum temperature reached by the gasses in a cycle
3) ##W_{net}^{mono}## and ##W_{net}^{bi}##
4) The efficiency of the heat engine for both gasses.
Relevant Equations
##\Delta U = W + Q##
I think I know how to solve the problem, but I've incurred into some problems with my computations.
Let's take for example the first question, in the monoatomic case.

The total heat absorbed by the gas coincides with the heat absorbed during ##A\rightarrow B##.
Since the pressure is constant ##Q_{AB}= nc_p \Delta T##. To find the heat I need to first find ##T_A## and ##T_B##. That's not a problem since The gas is ideal hence: $$T_A= \frac {P_A V_A} {nR}= \frac {(20) (5)} {(1)0.0821}= 1218K ; T_B= \frac {(20) (10)} {(1)0.0821}= 2436K $$
Therefore ##Q_{AB}= nc_p \Delta T= \frac 5 2 8.31 (2436 - 1218)= 25304J##.

The problem is that if I find ##\Delta U_{AB} = nc_v\Delta T= \frac 3 2 8.31 (2436-1218)\approx 15182J##, now ##\Delta U = Q + W## but that isn't the case since ##W_{AB}= -P_{AB} \Delta V= -20(10-5)= -100J##
##\Rightarrow 25304J -100J \neq 15182J##. I don't know where I'm making a mistake, but there must be one.
 
Physics news on Phys.org
  • #2
It might help to include units in your calculation.
##(20{\rm\ atm})(10 {\rm\ L} - 5 {\rm\ L})=100{\rm\ atm\cdot L}##
 
  • Like
Likes Thermofox
  • #3
robphy said:
What are your units of pressure and volume in your work calculation?
The numbers for the temperatures come out correctly when pressures in atm are converted to Pa, volumes in L converted to m-3 and 0.0821 (R) in the denominator is converted to 8.31 J/(mol K).

What I don't understand is if we have a heat engine with 1 mole of monatomic and 1 mole of diatomic gas mixed together or two separate heat engines with one mole of gas each.

robphy said:
It might help to include units in your calculation.
I think OP should do the calculations symbolically and then substitute. It will ease considerably the task of tracing what is going on and offer help.
 
  • Like
Likes Thermofox
  • #4
kuruman said:
two separate heat engines with one mole of gas each.
This.
robphy said:
It might help to include units in your calculation.
##(20{\rm\ atm})(10 {\rm\ L} - 5 {\rm\ L})=100{\rm\ atm\cdot L}##
kuruman said:
The numbers for the temperatures come out correctly when pressures in atm are converted to Pa, volumes in L converted to m-3 and 0.0821 (R) in the denominator is converted to 8.31 J/(mol K).
That's the mistake, I treated ##L\space atm## as ##J##

When I did ##W_{AB}= -P_{AB} (V_B - V_A)= -20atm(10-5)L\neq -100J##, but ##-100 atm\space L=101,325J(-100)\approx -10,132J##

Now ##\Delta U_{AB} = Q + W= 25,304J -10,132J = 15,172J##.

If I compare it with ##\Delta U= nc_v\Delta T_{AB}= 15,182J##, there is a difference of ##10J## but it is normal considering the approximations I did.
 
  • #5
Thermofox said:
The problem is that if I find ##\Delta U_{AB} = nc_v\Delta T= \frac 3 2 8.31 (2436-1218)\approx 15182J##, now ##\Delta U = Q + W## but that isn't the case since ##W_{AB}= -P_{AB} \Delta V= -20(10-5)= -100J##
##\Rightarrow 25304J -100J \neq 15182J##. I don't know where I'm making a mistake, but there must be one.
The problem is that you are substituting numbers when you should be using symbols.
For A→B
Note that since the volume doubles at constant pressure, ##T_B=2T_A##
Then
##\Delta U =\frac{3}{2}nR(2T_A-T_A)=\frac{3}{2}nRT_A=\frac{3}{2}p_AV_A.##
##Q=\frac{5}{2}nR(2T_A-T_A)=\frac{5}{2}nRT_A=\frac{5}{2}p_AV_A.##
##W=-p_AV_A.##
Clearly the first law is satisfied,$$\Delta U=Q+W\rightarrow\frac{3}{2}p_AV_A=\frac{5}{2}p_AV_A-p_AV_A.$$Now you substitute the given values for ##V_A## and ##p_A.## Be sure to convert to Pascals for pressure and cubic meters for volume.
 
  • Like
Likes Thermofox
  • #6
kuruman said:
The problem is that you are substituting numbers when you should be using symbols.
For A→B
Note that since the volume doubles at constant pressure, ##T_B=2T_A##
Then
##\Delta U =\frac{3}{2}nR(2T_A-T_A)=\frac{3}{2}nRT_A=\frac{3}{2}p_AV_A.##
##Q=\frac{5}{2}nR(2T_A-T_A)=\frac{5}{2}nRT_A=\frac{5}{2}p_AV_A.##
##W=-p_AV_A.##
Clearly the first law is satisfied,$$\Delta U=Q+W\rightarrow\frac{3}{2}p_AV_A=\frac{5}{2}p_AV_A-p_AV_A.$$Now you substitute the given values for ##V_A## and ##p_A.##
Okay, I was able to finish the problem. Thanks for the help!
 
  • Like
Likes kuruman

FAQ: Thermodynamic cyclic process - 2 isobaric and 2 adiabatic

What is a thermodynamic cyclic process?

A thermodynamic cyclic process is a series of thermodynamic transformations that return a system to its initial state. During this process, the system undergoes changes in pressure, volume, and temperature, but ultimately, the net change in internal energy and other state functions is zero, as the system returns to its original conditions.

What are isobaric and adiabatic processes?

An isobaric process is one that occurs at constant pressure, meaning that the pressure of the system remains unchanged while other variables like volume and temperature may change. An adiabatic process, on the other hand, is one in which no heat is exchanged with the surroundings; all changes in internal energy are due to work done on or by the system.

How does a cyclic process with 2 isobaric and 2 adiabatic steps work?

In a cyclic process with 2 isobaric and 2 adiabatic steps, the system undergoes two stages at constant pressure (isobaric) where heat is added or removed, followed by two adiabatic steps where the system expands or compresses without heat exchange. This combination allows for efficient energy transfer and can be used in heat engines to convert heat into work.

What are the applications of a thermodynamic cyclic process with isobaric and adiabatic steps?

This type of cyclic process is commonly found in various heat engines, such as the Carnot cycle and the Rankine cycle. These processes are fundamental in the design of power plants, refrigeration systems, and air conditioning units, where efficient energy conversion and temperature control are essential.

What is the significance of the first law of thermodynamics in cyclic processes?

The first law of thermodynamics, which states that energy cannot be created or destroyed, is crucial in cyclic processes. It ensures that the work done by the system plus the heat exchanged equals the change in internal energy. In a complete cycle, since the system returns to its initial state, the total change in internal energy is zero, meaning that the work done equals the heat added or removed during the process.

Similar threads

Internal energy won't add up to 0 in cyclic process
Replies
1
Views
487
Change of entropy in the Universe in a thermodynamic cycle
Replies
10
Views
2K
Closed cycle of an ideal gas
Replies
3
Views
720
Chemistry Entropy in isolated composite system for irreversible process
Replies
3
Views
836
Chemistry Calculation of thermodynamic variables for irreversible adiabatic process of ideal gas
Replies
9
Views
723
Thermodynamics: Cyclic Processes
Replies
2
Views
3K
Heat engine that uses a rubber band (Thermodynamics)
Replies
2
Views
2K
Violation of first/second law in a Cylic process
Replies
19
Views
2K
Thermodynamic properties - Isochronic process - First law
Replies
3
Views
2K
Chemistry How to explain the difference between recompression of ideal gas reversibly and irreversibly after initial irreversible expansion?
Replies
5
Views
641

Hot Threads

Recent Insights

Back
Top