Entropy of an adiabatic process

[luna02525]

Homework Statement

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.

Homework 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

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.

Thank you for your time reading this and possibly responding :D

 

[Andrew Mason]

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.

Start with the definition. For reversible processes: dS = dQ/T

$$\int_A^B dS = S_B - S_A = \Delta S = \int_A^B dQ/T$$

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

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: $W = P\Delta V$. 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.

AM

 

[baba89]

Hello! I can help you with your questions regarding entropy of an adiabatic process.

Firstly, for part a), you have correctly calculated the values of q, w, ΔU, and ΔH. However, for the entropy change (ΔS), it is important to note that in an adiabatic process, there is no heat transfer (q=0). Therefore, the change in entropy for the system (ΔS_sys) will also be zero. However, for the surroundings, there is a change in entropy (ΔS_surr) due to the work done on the surroundings by the gas. This can be calculated using the equation ΔS_surr = -q/T = -w/T. Plugging in the values of w and T, you should get a value of -3.01 J/K mol for ΔS_surr. Thus, the total change in entropy (ΔS_tot) will be the sum of ΔS_sys and ΔS_surr, which is -3.01 J/K mol.

For part b), as the process is not reversible, we cannot use the equation for reversible adiabatic expansion. Instead, we can use the equation you mentioned in your attempt, which is ΔU = w = -P_extΔV. Since the external pressure (P_ext) is constant at 0.5 atm, we can use this equation to find the final volume (V_2) and temperature (T_2). Once we have these values, we can construct a reversible path with two steps, where the gas expands at constant temperature (T_2) and then at constant pressure (P_ext). This will allow us to calculate the values of q, w, ΔU, ΔH, and ΔS for this path.

I hope this helps! Let me know if you have any further questions.