1. Sep 28, 2008

luna02525

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, $$\Delta$$U, $$\Delta$$H, $$\Delta$$S, $$\Delta$$Ssurr, and $$\Delta$$Stot for each path where the data permit. Take C_v,m = 3/2 R (HINT: For part b: use the fact that $$\Delta$$U = w = nC_v$$\Delta$$T = -Pext$$\Delta$$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
$$\Delta$$U = w = int C_p dT
$$\Delta$$H = int C_v dT
$$\Delta$$S = $$\Delta$$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 $$\Delta$$U = w = -905.931 J/mol and $$\Delta$$H = -1509.884 J/mol

For part a I am unsure of how to find entropy of an adiabatic system expanding reversibly...my ideas:
$$\Delta$$S = 0 $$\Delta$$S_sys = C_p*ln(T_2/T_1) = 5.76 J/k mol, and therefore $$\Delta$$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.

2. Sep 28, 2008

Andrew Mason

$$\int_A^B dS = S_B - S_A = \Delta S = \int_A^B dQ/T$$
If dQ = 0 for the gas, then what is dQ for the surroundings? What, then is dS and, therefore, $\Delta S_{surr} \text{ and } \Delta S_{gas}$?
For expansion at constant pressure, it is a simple matter to determine the work done: $W = P\Delta V$. Use the first law, dQ = dU + dW to determine what $\Delta U$ 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.