(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Homework Help: Entropy of an adiabatic process

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