Adiabatic process, calculating final T, enthelpy etc

In summary, the gas expands against a constant external pressure, but is not allowed to fully expand. The volume change occurs at a constant applied external pressure, and the work done by the gas on its surroundings is determined by combining the equation for the first law of thermodynamics with the ideal gas law.
  • #1
PhilJones
9
1

Homework Statement


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


Initial state of gas
Ti = 353K
Pi = 250000Pa
2.5mols of gas
Cv = 12.47Jmol-1

Final pressure = 125000Pa

Homework Equations


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

The Attempt at a Solution


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

so Tfinal = 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 = PVγ * (V21-γ - V11-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.
PVγ = 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
 
Last edited:
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  • #2
Would it be possible for you to provide the exact problem statement?
 
  • #3
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
 
  • #4
edit: sorry posted same comment twice..
 
  • #5
PhilJones said:
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}##.
 
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  • #6
Chestermiller said:
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 = nCVΔT = -W = -Pext(Vf - Vi)

so nCv(Tf - Ti) = -Pext(Vf - Vi)

Would that be correct?
 
  • #7
PhilJones said:
Thanks for explaining what's actually physically going on, I had no idea.

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

ΔU = nCVΔT = -W = -Pext(Vf - Vi)

so nCv(Tf - Ti) = -Pext(Vf - Vi)

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?
 
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  • #8
Chestermiller said:
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,

Vi = nRTi/Pi = (2.5*8.314*325)/(250000) ≈ 0.027 m3
and
Vf = nRTf/Pf = (2.5*8.314*Tf)/(125000) ≈ 1.66*10-4*Tf

and so using nCv(Tf - Ti) = -Pext(Vf - Vi)

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

31.18 Tf + 16.6 Tf = 11005 + 2700

≈47.8 Tf = 13705 => Tf ≈ 289K

so barring mistakes
a) Tf = 289K

b) Q = 0

c) W, work = PextΔV

Vf = nRTf/Pf = 2.5*8.314*289/125000 ≈ 0.048 m3
so W = PextΔV = 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
 
  • #9
PhilJones said:
Thanks so much for the help! Sorry, awkwardness of timezones means this late reply.

OK so,

Vi = nRTi/Pi = (2.5*8.314*325)/(250000) ≈ 0.027 m3
and
Vf = nRTf/Pf = (2.5*8.314*Tf)/(125000) ≈ 1.66*10-4*Tf

and so using nCv(Tf - Ti) = -Pext(Vf - Vi)

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

31.18 Tf + 16.6 Tf = 11005 + 2700

≈47.8 Tf = 13705 => Tf ≈ 289K

so barring mistakes
a) Tf = 289K

b) Q = 0

c) W, work = PextΔV

Vf = nRTf/Pf = 2.5*8.314*289/125000 ≈ 0.048 m3
so W = PextΔV = 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.
 
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  • #10
Chestermiller said:
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 TVγ-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
 
  • #11
PhilJones said:
Ah yes they do, thanks a lot much appretiated.

Out of interest, is TVγ-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.
 
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  • #12
Chestermiller said:
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.
 

Related to Adiabatic process, calculating final T, enthelpy etc

1. What is an adiabatic process?

An adiabatic process is a thermodynamic process in which there is no heat exchange between the system and its surroundings. This means that the system is isolated and there is no transfer of heat or matter.

2. How do you calculate the final temperature in an adiabatic process?

The final temperature in an adiabatic process can be calculated using the equation T2 = T1 * (P2/P1)^(γ-1), where T1 and T2 are the initial and final temperatures, P1 and P2 are the initial and final pressures, and γ is the heat capacity ratio of the gas.

3. What is enthalpy and how is it related to adiabatic processes?

Enthalpy is a thermodynamic property that represents the total energy of a system. In an adiabatic process, where there is no heat exchange, the change in enthalpy (ΔH) is equal to the change in internal energy (ΔU).

4. How do you determine if a process is adiabatic?

A process is considered adiabatic if there is no heat exchange between the system and its surroundings. This can be determined by observing the system for any heat transfer, such as changes in temperature or phase changes.

5. What are some real-life examples of adiabatic processes?

Some examples of adiabatic processes in everyday life include the compression of gas in a car engine, the expansion of gas in a refrigerator, and the heating of air by a hair dryer. These processes all occur without any heat exchange between the system and its surroundings.

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