A sample consisting of 1 mol of He gas is initially at 240 kPa and 325 K. It undergoes reversible adiabatic expansion until its pressure reaches 150 kPa. Using equipartition theorem, find the Cvm, Cpm and y. Calculate the final volume, final temperature and the work done? (Assume it is perfect gas)
A sample consisting of 1 mol of He gas is initially at 240 kPa and 325 K. It undergoes reversible adiabatic expansion until its pressure reaches 150 kPa. Using equipartition theorem, find the Cvm, Cpm and y. Calculate the final volume, final temperature and the work done? (Assume it is perfect gas)
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:7. When a perfect gas is expanded adiabatically its intemal energy is reduced by the
expansion work. This can be shown as:
AU = q + w and since it is adiabatic process: q = 0. Therefore:
AU = wad = -Pezt .dV
where AU of perfect gas is independent of volume molecules occupy, which means that
dU = Cy.dT. If the process is reversible pezt = Peys = P; which is for perfect gases:
nRT
-. Both sides can be combined to find:
p=
V
-nRT
-dV. This equation can be rearranged to:
Cv. dT =
dT
Cy.
T
V
-nR
dv. The temperature of the gas changes by this equation when the volume
changes. We can integrate the equation:
dT
AP
Cy
-nR
T
The result of the equation will be:
(Tj
Cy. In
(주)
-nR. In
Because In(x'y) =-In(y/x), this expression rearranges to:
Cy
.In
T;
) --)
= In
With c = Cy/nR we obtain (because In x = a ln x):
In
= In
1/e
T; =T;
where c = Cy/nR= Cvm R
This equation is equivalent to: V,T; = V;T; (Adiabatic reversible perfect gas).
For perfect gases we can write:
P:V
P;V; Tj
If the initial and final temperatures were replaced in the previous equation:
Transcribed Image Text:1/c
PfV;
+1
Pi
= 1
Pf
We now use the result from Topic 2B that C,m - Crm =R to note that
Cv m +R Cp,m
Cv m
1+ Cvm/R
Cv=/R
1. 1+c
+:
Cv m
The equation is now updated to:
Pi
= 1
The equation becomes:
P:V;" = PfV;"
This is reversible adiabatic expansion of perfect gas. For a monoatomic gas Cym = 32R
and Cpm = 5,R. So, the adiabatic exponent is y= 5/3.
A sample consisting of 1 mol of He gas is initially at 240 kPa and 325 K. It undergoes
reversible adiabatic expansion until its pressure reaches 150 kPa. Using equipartition theorem,
find the Cv,m, Cp,m and y. Calculate the final volume, final temperature and the work done?
(Assume it is perfect gas)
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