finding work, heat, and entropy in a heat engine


by rubenhero
Tags: thermodymanics
rubenhero
rubenhero is offline
#1
Dec14-11, 10:12 AM
P: 42
1. The problem statement, all variables and given/known data
An engine operates with n = 1.94 moles of a monatomic ideal gas as its working substance, starting at volume VA = 21.4 L and pressure PA = 2.77 atm. The cycle consists of:

- A to B: an isobaric expansion until its volume is VB = 50.1 L
- B to C: an isochoric (constant volume) process that lowers the pressure to PC = 1.184 atm
- C to A: an isothermal (constant temperature) process that returns the gas to its original pressure and volume

HINT: Draw a P-V diagram of this process.

a) Find TT, the absolute temperature of the isothermal process.
b) Find WT, the work done on the gas during the isothermal process.
c) Find ΔST, the change in entropy of the gas during the isothermal process.
d) Find Wtot, the work done in the entire cycle.
e) Find Qtot, the heat that is exchanged (in or out) during the entire cycle.


2. Relevant equations
PV=nRT, Wdone on a gas = nRTln(Vo/Vf),
ΔS = Qrev/T, Qp = CpT, Cp = 5/2nR,
Qv = CvΔT, Cv = 3/2nR


3. The attempt at a solution
necessary conversions:
1 atm = 1.013e5 Pa
2.77 atm = 280601 Pa
1.184 atm = 119939.2 Pa

1L = 1/1000 m^3
21.4 L = .0214 m^3
50.1 L = .051 m^3

2 a) Find TT, the absolute temperature of the isothermal process.
PV = nRT
PV/nR = T
(280601Pa * .0214m^3)/ (1.94 moles * 8.314 J/mol-K) = T
372.2984582K = T
372 K = T [webassign marked as correct] 3 significant digits

b) Find WT, the work done on the gas during the isothermal process.

Won the gas = nRT ln(Vo/Vf)
= 1.94 moles * 8.314 J/mol-K * 372 K ln(.051m^3/.0214m^3)
= 5214.830072 J
Won the gas = 5.21e3 J [webassign marked as correct] 3 significant digits

c) Find ΔST, the change in entropy of the gas during the isothermal process.
change in IE is 0 since it is an isothermal process, work is positive because the gas was compressed and work was done on the gas
change IE = Q + W
0 = Q + 5214.830072 J
-5214.830072 J = Q
dS = dQrev/T
= -5214.830072 J/ 372.2984582 K
= -14.0071224 J/K
dS = -14.0 J/K [webassign marked as wrong] 3 significant digits

d) Find Wtot, the work done in the entire cycle.
plan : find work done in all parts of the cycle and get sum

A-B is an isobaric process so W = -PdV
= -280601 Pa * (.051m^3 - .0214m^3)
= -8365.7896 J
B-C is an isochoric process so W = 0
C-A is an isothermal process (taken from part 2b) W = 5214.830072 J

Wtot = WA-B + WB-C + WC-A = -8365.7896 J + 0J + 5214.830072 J
Wtot = -3090.959528 J
Wtot = -3.09e3 J [ webassign marked as wrong] 3 significant digits

e) Find Qtot, the heat that is exchanged (in or out) during the entire cycle.

plan : find heat in all parts of the cycle and get sum

A-B is an isobaric process so Q = CpdT
= 5/2 n R dT

we need the different temperature at points A and B to get dT:
@ A : PV/nR = T
= (280601Pa * .0214m^3)/ (1.94 moles * 8.314 J/mol-K) = T
= 372 K
@B : PV/nR = T
= (280601Pa * .051m^3)/ (1.94 moles * 8.314 J/mol-K) = T
= 887.253335 K
dT = TB - TA = 887.253335 K - 372.2984582 K = 515 K

so now QA-B = 5/2 n R dT
= 5/2 * 1.94 moles * 8.314 J/mol-K * 515 K
= 20764.474 J positive because heat was added to the system

B-C is an isochoric process so Qv = CvdT
= 3/2 n R dT
we need the different temperature at points B and C to get dT:
@B : PV/nR = T
= (280601Pa * .051m^3)/ (1.94 moles * 8.314 J/mol-K) = T
= 887.253335 K
@C : PV/nR = T
= (119939.2Pa * .051m^3) / (1.94 moles * 8.314 J/mol-K) = T
= 379.2447468 K
dT = TB - Tc = 887.253335 K - 379.2447468 K
= 508.0085882 K
= 508 K
so Q = 3/2 n R dT
= 3/2 1.94 moles * 8.314 J/mol-K * 508 K
= -12290.6277 J negative because heat was removed
C-A is an isothermal process and we already figured out Q is -5214.830072 J from part 2b

so finally Qtot = QA-B + QB-C + QC-A

= 20764.474 J + -12290.6277 J + -5214.830072 J
= 3259.016228 J
= 3.25e3 J [webassign marked as wrong] 3 significant figures

thank you in advance for any input as to why my work might be wrong.
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ehild
ehild is offline
#2
Dec14-11, 10:57 AM
HW Helper
Thanks
P: 9,818
Quote Quote by rubenhero View Post
3. The attempt at a solution
necessary conversions:


1L = 1/1000 m^3
21.4 L = .0214 m^3
50.1 L = .051 m^3
VB=0.0501 m^3.
Recalculate the isotherm work with the correct volume. Also try to use more significant digits in T.

ehild
rubenhero
rubenhero is offline
#3
Dec14-11, 11:23 AM
P: 42
Thank you ehild, I redid all the work and got the correct answers.


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