What is the Efficiency of an Ideal Monatomic Gas Cycle?

  • Thread starter Thread starter Vandella
  • Start date Start date
  • Tags Tags
    Cycle Gas
AI Thread Summary
The discussion focuses on calculating the efficiency of an ideal monatomic gas cycle involving a 1.00 mol sample. Key calculations include the net work done by the gas, which was determined to be 1.68 kJ after evaluating the work during the isothermal expansion and subsequent processes. The temperature at point A was confirmed to be 1822 K, and the heat energy supplied during the expansion was noted to be approximately 7.73 kJ. However, participants highlighted the lack of a diagram or complete information about the rest of the cycle, which is necessary for finalizing the calculations. Overall, the conversation emphasizes the importance of having complete cycle details for accurate thermodynamic analysis.
Vandella
Messages
25
Reaction score
0

Homework Statement



A 1.00 mol sample of an ideal monatomic gas is taken through the cycle shown in the figure. The process A B is a reversible isothermal expansion where PA = 5.0 atm, PB = 3.0 atm, VA = 30.0 L, and VB = 50.0 L.

(a) Calculate the net work done by the gas.

(b) Calculate the energy added to the gas by the heat.

(c) Calculate the energy exhausted from the gas by heat.

(d) Calculate the efficiency of the cycle.


Homework Equations





The Attempt at a Solution


started by calculating tamp at point A using (pressureA*VolumeA)/number of moles*gas constant, that came to 1822.2K
then used work=nRT ln(VA/VB) think that came to -7.74kj
next step W2=PB(VB-VC) which i got 6.06kj as answer

so net work done 7.74-6.06=1.68

for the following parts used Q2=n(5/2)R(Tc-Tb) to show energy exhausted
and Q3=n(3/2)R(Tb-Tc) combined this with W1=Q1 to get energy added

at some point i have made an error please help
 
Physics news on Phys.org
I agree with the temp you calculated for point A. From A to B is isothermal so the temp along this expansion is 1822K.
But... there is no diagram or information about the rest of the cycle!
 
I also got the work done BY THE GAS in going from A to B to be 7.73kJ.
This is the amount of heat energy which must be supplied to the gas going from A to B.
Need the other details of the cycle to complete the calculations.
You are obviously on the right tracks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top