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Entropy of an adiabatic process

by luna02525
Tags: adiabatic, entropy, process
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luna02525
#1
Sep28-08, 01:25 PM
P: 13
1. The problem statement, all variables and given/known data

A sample of 1.00 mol of an ideal gas at 300 K and 101325 Pa is expanded adiabatically in two ways: a) reversibly to .5 atm and b) against a constant external pressure of .5 atm. Determine the values of q, w, [tex]\Delta[/tex]U, [tex]\Delta[/tex]H, [tex]\Delta[/tex]S, [tex]\Delta[/tex]Ssurr, and [tex]\Delta[/tex]Stot for each path where the data permit. Take C_v,m = 3/2 R (HINT: For part b: use the fact that [tex]\Delta[/tex]U = w = nC_v[tex]\Delta[/tex]T = -Pext[tex]\Delta[/tex]V to find the final temperature, and then construct a reversible path of 2 steps, where P and T change only one at a time.



2. Relevant equations

q = 0 in all adiabatic processes
C_p,m = 5/2*R
C_v,m = 3/2*R
[tex]\Delta[/tex]U = w = int C_p dT
[tex]\Delta[/tex]H = int C_v dT
[tex]\Delta[/tex]S = [tex]\Delta[/tex]H/T = intC_p/T dT

for part a) reversible adiabatic: T_2/T_1 = (P_2/P_1)^[(gamma-1)/gamma] where gamma = C_p/C_v = 5/3


3. The attempt at a solution

for part a)

I used the equation for an ideal gas in reversible adiabatic expansion to find T_2

T_2/300 K = (50662.5 Pa/ 101325 Pa)^(2/5)
T_2 = 227.357 K

from there I got [tex]\Delta[/tex]U = w = -905.931 J/mol and [tex]\Delta[/tex]H = -1509.884 J/mol

For part a I am unsure of how to find entropy of an adiabatic system expanding reversibly...my ideas:
[tex]\Delta[/tex]S = 0 [tex]\Delta[/tex]S_sys = C_p*ln(T_2/T_1) = 5.76 J/k mol, and therefore [tex]\Delta[/tex]S_tot = 5.76 J/k mol but this seems like an unreasonable answer.

And for part b I am sort of lost for where to start.

1) I know all of the above stated equations and unknowns. And that hint to find T_2 but I don't know how to find it! I have plugged in Starting P, n, and T's into the ideal gas equation but this only yields V_1 to plug into my work equation to find T_2.

Thank you for your time reading this and possibly responding :D
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Andrew Mason
#2
Sep28-08, 05:00 PM
Sci Advisor
HW Helper
P: 6,684
For part a I am unsure of how to find entropy of an adiabatic system expanding reversibly...my ideas:
[tex]\Delta[/tex]S = 0 [tex]\Delta[/tex]S_sys = C_p*ln(T_2/T_1) = 5.76 J/k mol, and therefore [tex]\Delta[/tex]S_tot = 5.76 J/k mol but this seems like an unreasonable answer.
Start with the definition. For reversible processes: dS = dQ/T

[tex]\int_A^B dS = S_B - S_A = \Delta S = \int_A^B dQ/T[/tex]

If dQ = 0 for the gas, then what is dQ for the surroundings? What, then is dS and, therefore, [itex]\Delta S_{surr} \text{ and } \Delta S_{gas}[/itex]?

And for part b I am sort of lost for where to start.
This is a little trickier. dS = dQ/T only for reversible processes.

For expansion at constant pressure, it is a simple matter to determine the work done: [itex]W = P\Delta V[/itex]. Use the first law, dQ = dU + dW to determine what [itex]\Delta U[/itex] will be when expansion is complete and this will give you the temperature of the gas. Since entropy is a state function, the change in entropy in the non-reversible process will be equal to the entropy change resulting from a reversible path between those two states.

AM


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