
#1
Sep2808, 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)^[(gamma1)/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 



#2
Sep2808, 05:00 PM

Sci Advisor
HW Helper
P: 6,562

[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]? 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 nonreversible process will be equal to the entropy change resulting from a reversible path between those two states. AM 


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