Adiabatic process, calculating final T, enthelpy etc

[PhilJones]

Homework Statement

[/B]
Find the final temperature, Q, ΔU, ΔH given the following


Initial state of gas
T_i = 353K
P_i = 250000Pa
2.5mols of gas
C_v = 12.47Jmol^-1

Final pressure = 125000Pa

Homework Equations

PV = nRT
W = -PΔV
ΔH = ΔU + Δ(PV)
PV^γ = constant

The Attempt at a Solution

C_v / R ≈3/2 so monatomic => γ = 5/2
PV=nRT => V1 = (2.5)(8.314)(325)/(250000) = 0.027 m³
using PV^γ = constant => V2 = (P1 / P2 *V^γ )^1/γ = (2.5/1.25)^2/5 * (0.027)= 0.0356 m³

so T_final = P2 *V2 /(n*R) = [(125000*0.0356)/(2.5*8.314)] = 214K
Q = 0 adiabatic
W = ? nCv ΔT = 2.5*12.47*1(-139) = -4333J?
I get a different answer if I use W = PV^γ * (V2^1-γ - V1^1-γ)/(1-γ) = 1527
ΔU = W = -4333J
H = H = ΔU + Δ(PV), should this be = ΔU + nRΔT or = ΔU + PΔV + VΔP? They don't give me the same answer

nRΔT = 2785J, PΔV + VΔP = 5525J but I used P = 125000Pa and V = 0.0356, i don't know if this is the right formula let alone the right parameters to enter.

I didnt even use the fact that the external pressure was 1bar either.

All round confused with this part of my thermal course and just applied random formulas so I think I botched it.
PV^γ = constant is something from another course I did and wasn't in my notes so is it the correct route or maybe an there's an easier method? Any confirmation/corrections would be much appreciated.

Edit: made numerical errors that I corrected

 

[Chestermiller]

Would it be possible for you to provide the exact problem statement?

 

[PhilJones]

Sure,

this is an image of it, it was a past exam question. http://i.imgur.com/KT8XOOx.png
sorry I definitely wasn't clear enough in my initial post, I see that now

 

[PhilJones]

edit: sorry posted same comment twice..

 

[Chestermiller]

Sure,

this is an image of it, it was a past exam question. http://i.imgur.com/KT8XOOx.png
sorry I definitely wasn't clear enough in my initial post, I see that now

OK. Thanks. That helps.

There is a little bit of trickiness to this problem. Here's what happens:

To start with, the (massless, frictionless) piston is sitting on top of the gas. The pressure outside the cylinder is 1 bar, and the pressure inside the cylinder is 2.5 bars. So the piston has to be held in place. At time zero, the piston is released, and the gas expands against a constant external pressure of $P_{ext}=1$ bar. But, the gas is not allowed to fully expand to match the external pressure of 1 bar. Instead, at some final volume $V_f$ the piston is again constrained. Once this happens, the gas re-equilibrates uniformly to a new final pressure of 1.5 bars.

So the volume change occurs at a constant applied external pressure of 1 bar (which matches the local gas pressure on the gas side of the piston face). So the work done by the gas on its surroundings during this process is $W=P_{ext}(V_f-V_i)$

$V_i$ is known from the ideal gas law, but, since the final temperature $T_f$ is unknown, the final pressure $V_f$ is also unknown. So there are two unknowns in this problem. These two unknowns must be determined by combining the equation for the first law of thermodynamics with the ideal gas law. Please (algegraically) write down the equation for the first law of thermodynamics that would apply to this system, in terms of n, $C_v$, $T_i$, $T_f$, $V_i$, $V_f$, and $P_{ext}$.

 

[PhilJones]

OK. Thanks. That helps.

There is a little bit of trickiness to this problem. Here's what happens:

To start with, the (massless, frictionless) piston is sitting on top of the gas. The pressure outside the cylinder is 1 bar, and the pressure inside the cylinder is 2.5 bars. So the piston has to be held in place. At time zero, the piston is released, and the gas expands against a constant external pressure of $P_{ext}=1$ bar. But, the gas is not allowed to fully expand to match the external pressure of 1 bar. Instead, at some final volume $V_f$ the piston is again constrained. Once this happens, the gas re-equilibrates uniformly to a new final pressure of 1.5 bars.

So the volume change occurs at a constant applied external pressure of 1 bar (which matches the local gas pressure on the gas side of the piston face). So the work done by the gas on its surroundings during this process is $W=P_{ext}(V_f-V_i)$

$V_i$ is known from the ideal gas law, but, since the final temperature $T_f$ is unknown, the final pressure $V_f$ is also unknown. So there are two unknowns in this problem. These two unknowns must be determined by combining the equation for the first law of thermodynamics with the ideal gas law. Please (algegraically) write down the equation for the first law of thermodynamics that would apply to this system, in terms of n, $C_v$, $T_i$, $T_f$, $V_i$, $V_f$, and $P_{ext}$.

Thanks for explaining what's actually physically going on, I had no idea.

OK so,
ΔU = Q - W, adiabatic => Q = 0
ΔU = -W

ΔU = nC_VΔT = -W = -P_ext(V_f - V_i)

so nC_v(T_f - T_i) = -P_ext(V_f - V_i)

Would that be correct?

 

[Chestermiller]

Thanks for explaining what's actually physically going on, I had no idea.

OK so,
ΔU = Q - W, adiabatic => Q = 0
ΔU = -W

ΔU = nC_VΔT = -W = -P_ext(V_f - V_i)

so nC_v(T_f - T_i) = -P_ext(V_f - V_i)

Would that be correct?

Yes. Good job. Now for applying the ideal gas law.

Algebraically, from the ideal gas law, what is the initial volume $V_i$ in terms of $T_i$, $P_i$, and n?
Algebraically, from the ideal gas law, what is the final volume $V_f$ in terms of $T_f$, $P_f$, and n (where $P_f$ is the 1.25 bars)?
If you algebraically substitute these volume equations into your result from the ideal gas law, algebraically, what do you get?

 

[PhilJones]

Yes. Good job. Now for applying the ideal gas law.

Algebraically, from the ideal gas law, what is the initial volume $V_i$ in terms of $T_i$, $P_i$, and n?
Algebraically, from the ideal gas law, what is the final volume $V_f$ in terms of $T_f$, $P_f$, and n (where $P_f$ is the 1.25 bars)?
If you algebraically substitute these volume equations into your result from the ideal gas law, algebraically, what do you get?

Thanks so much for the help! Sorry, awkwardness of timezones means this late reply.

OK so,

V_i = nRT_i/P_i = (2.5*8.314*325)/(250000) ≈ 0.027 m³
and
V_f = nRT_f/P_f = (2.5*8.314*T_f)/(125000) ≈ 1.66*10^-4*T_f

and so using nC_v(T_f - T_i) = -P_ext(V_f - V_i)

2.5*12.47*(T_f - 353) = -100000(1.66*10^-4*T_f - 0.027)

31.18 T_f + 16.6 T_f = 11005 + 2700

≈47.8 T_f = 13705 => T_f ≈ 289K

so barring mistakes
a) T_f = 289K

b) Q = 0

c) W, work = P_extΔV

V_f = nRT_f/P_f = 2.5*8.314*289/125000 ≈ 0.048 m³
so W = P_extΔV = 100000*(0.048 - 0.027) = 2100 J

d) ΔU = just minus the work

e) ΔH = ΔU + Δ(PV) I'm unsure how to use this formula in terms of the Δ(PV), should I use product rule or just nRΔT

 

[Chestermiller]

Thanks so much for the help! Sorry, awkwardness of timezones means this late reply.

OK so,

V_i = nRT_i/P_i = (2.5*8.314*325)/(250000) ≈ 0.027 m³
and
V_f = nRT_f/P_f = (2.5*8.314*T_f)/(125000) ≈ 1.66*10^-4*T_f

and so using nC_v(T_f - T_i) = -P_ext(V_f - V_i)

2.5*12.47*(T_f - 353) = -100000(1.66*10^-4*T_f - 0.027)

31.18 T_f + 16.6 T_f = 11005 + 2700

≈47.8 T_f = 13705 => T_f ≈ 289K

so barring mistakes
a) T_f = 289K

b) Q = 0

c) W, work = P_extΔV

V_f = nRT_f/P_f = 2.5*8.314*289/125000 ≈ 0.048 m³
so W = P_extΔV = 100000*(0.048 - 0.027) = 2100 J

d) ΔU = just minus the work

e) ΔH = ΔU + Δ(PV) I'm unsure how to use this formula in terms of the Δ(PV), should I use product rule or just nRΔT

You can use $(P_fV_f-P_iV_i)$ of $nR\Delta T$ or $\Delta H=nC_p\Delta T$. They should all give the same result.

 

[PhilJones]

You can use $(P_fV_f-P_iV_i)$ of $nR\Delta T$ or $\Delta H=nC_p\Delta T$. They should all give the same result.

Ah yes they do, thanks a lot much appretiated.

Out of interest, is TV^γ-1 = constant applicable to this question? When I tried just there for T1, T2 and V1 to find V2 it didnt work, I got V₂ = 0.0365 m³

 

[Chestermiller]

Ah yes they do, thanks a lot much appretiated.

Out of interest, is TV^γ-1 = constant applicable to this question? When I tried just there for T1, T2 and V1 to find V2 it didnt work, I got V₂ = 0.0365 m³

That equation is only valid for an adiabatic reversible process. This uncontrolled expansion against a constant pressure is a spontaneous irreversible process.

 

[PhilJones]

That equation is only valid for an adiabatic reversible process. This uncontrolled expansion against a constant pressure is a spontaneous irreversible process.

Thanks for helping my understanding, it's all a lot clearer now. Cheers for your time.