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« on: February 17, 2006, 11:22:54 AM »
Hi Geniuses,

Just need some help to check whether i have used the right calculations to work out a problem.

the question states:

Calculate the final temperature of a sample of carbon dioxide (assume ideal gas) of mass 16g that is expanded reversibly and adiabatically from 500mL to 2.00L. The constant pressure heat capacity of CO2 is 37.11 J k-1 mol-1 .

Here is my working:

1.) the process is adiabatic and reversible, therefore, (heat)q=0,
(work) w=0
and internal energy change delta U=0.

2.) entropy change, delta S = qrev/T= 0

So, if the pressure of the system is held constant (delta P=0) as indicated by the constant pressure heat capacity value given in the question and the moles (n) are also held constant, then the total entropy change of the system can be calculted by using the following equation:

delta Ssys= n*Cpln(Tfinal / Tinitial) + n*R ln(Vfinal/Vinitial)

Now if delta S is zero for a reversible adiabatic process, the above equation is equal to zero, therefore, the final temperature can be calculated by solving this equation. ( can't it )

Looking at the values given in the question:

n= 16g*(1mol CO2 / 44.0098g) = 0.36355589mol CO2

Vinitial=0.5L

Vfinal=2.0L

Tinitial=298.15K

Cp=37.11 J K-1 mol-1

and inserting into the equation:

[(0.36355589mol)(37.11 J K-1 mol-1) ln (Tfinal / 298.15K)] + [(0.36355589mol)(8.314472 J K mol) ln (2.0L / 0.5L)] = 0

Which simplifies to:

(13.49155908 J K-1) ln (Tfinal / 298.15K) + 4.190456309 J K = 0

so:

(13.49155908 J K-1) ln (Tfinal / 298.15K) = -4.190456309 J K

Rearranging:

ln (Tfinal / 298.15K) = -4.190456309 J K / 13.49155908 J K-1

so:

ln (Tfinal / 298.15K) = -0.310598373

then:

(Tfinal / 298.15K) = e-0.310598373

And finally:

Tfinal = (e-0.310598373)*(298.15K)

= 218.5463986K

Rounding off:

Tfinal = 218.55K

If anyone can confirm that this was the right method to use i would appreciate it greatly, is there a simpler way to calculate final temperature in a reversible adiabatic process?

cheers,

The only stupid question is a question not asked.

#### Donaldson Tan

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« Reply #1 on: February 17, 2006, 06:01:29 PM »
For perfect gas undergoing a reversible and adiabatic process. the perfect gas can be described by the follow equation: PVgamma = constant where gamma = Cp/Cv (Cv: heat capacity at constant volume)

for perfect gas, Cp - Cv = R=> Cv = Cp - R

gamma = Cp/Cv = Cp/(Cp-R) = 37.11/(37.11-8.314) = 1.288

since PVgamma = constant then
P1V1gamma = P2V2gamma

let (1) be the initial state
let (2) be the final state

Since we know the initial temperature and volume, and the final volume, then we can express PVgamma in terms of V and T by using the perfect gas equation PV = nRT

PV = nRT => P = nRT/V
PVgamma = (nRT/V)Vgamma = nRT.Vgamma - 1

P1V1gamma = P2V2gamma
nRT1.V1gamma - 1 = nRT2.V2gamma - 1
nRT1.V1gamma - 1 = nRT2.V2gamma - 1
T1.V1gamma - 1 = T2.V2gamma - 1

Tfinal = T2 = T1.(V1/V2)gamma - 1 = 298.15.(0.5/2.0)1.288 - 1 = 200K
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### Donaldson Tan

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« Reply #2 on: February 17, 2006, 06:32:47 PM »
1.) the process is adiabatic and reversible, therefore, (heat)q=0,
(work) w=0
and internal energy change delta U=0.
This is dodgy: adiabatic means Q = 0. It does not imply that W = dU = 0

So, if the pressure of the system is held constant (delta P=0) as indicated by the constant pressure heat capacity value given in the question and the moles (n) are also held constant, then the total entropy change of the system can be calculted by using the following equation ...
If pressure is constant, then you can use Charles' Law directly to find the final temperature. You need not use Cp at all to final the final temperature.

However, there is no grounds in assuming constant pressure. Below is my proof why constant pressure should be assumed:
Since this is an adiabatic process, then dU = W = -p.dV
dU = Cv.dT = -p.dV
for Perfect Gas, Cp - Cv = R => Cv = Cp - R
(Cp - R)dT = -p.dV
Cp.dT - R.dT = -p.dV
differentiating the perfect gas equation,
d(pV) = d(RT) => p.dV + V.dp = R.dT
Cp.dT - (p.dV + V.dp) = -p.dV
Cp.dT - V.dp = 0 => Cp.dT = V.dp

For reversible adiabatic process involving a perfect gas,
Unless dT = 0, then dP = 0. This means if the reversible adiabatic process is isobaric, then it must be isothermal. However, from the perfect gas equation (PV = nRT), it is evident that volume of a fixed mass of gas is constant if its pressure and temperature are held constant.

Since this question requires the gas to expand, then pressure and temperature cannot be constant. Hence, the assumption for constant pressure is invalid
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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« Reply #3 on: February 18, 2006, 03:13:40 PM »
Thanks for the guidence geodome, much appreciated

I found a different way ( slightly) of doing that question and got the same result you did.

_
R ln (V1/V2) = Cv ln ( T2/T1)
_
T2 = e(R ln (V1/V2))/Cv = 199.82K

cheers,

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#### Donaldson Tan

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« Reply #4 on: February 19, 2006, 07:24:04 AM »
R ln (V1/V2) = Cv ln ( T2/T1)

Your method is a sub-derivation from my proof.

Since we know that dU = - p.dV and dU = Cv.dT, then
Cv.dT = -p.dV = - (RT/V).dV
(Cv/T).dT = -(R/V)dV
integrate from T1 to T2(Cv/T).dT = integrate from V1 to V2 -(R/V)dV
=> Cv.ln(T1/T2) = -R.ln(V1/V2) = R.ln(V2/V1)

However, the PVgamma = constant is a more useful equation.

It is valid for a perfect gas in both open and close system, undergoing an isoentropic process.
« Last Edit: February 19, 2006, 07:24:33 AM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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« Reply #5 on: February 19, 2006, 09:03:35 AM »
Thanks geodome
The only stupid question is a question not asked.

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« Reply #6 on: February 19, 2006, 11:58:03 AM »
This is dodgy: adiabatic means Q = 0. It does not imply that W = dU = 0If pressure is constant, then you can use Charles' Law directly to find the final temperature. You need not use Cp at all to final the final temperature.

However, there is no grounds in assuming constant pressure. Below is my proof why constant pressure should be assumed:
Since this is an adiabatic process, then dU = W = -p.dV
dU = Cv.dT = -p.dV
for Perfect Gas, Cp - Cv = R => Cv = Cp - R
(Cp - R)dT = -p.dV
Cp.dT - R.dT = -p.dV
differentiating the perfect gas equation,
d(pV) = d(RT) => p.dV + V.dp = R.dT
Cp.dT - (p.dV + V.dp) = -p.dV
Cp.dT - V.dp = 0 => Cp.dT = V.dp

For reversible adiabatic process involving a perfect gas,
Unless dT = 0, then dP = 0. This means if the reversible adiabatic process is isobaric, then it must be isothermal. However, from the perfect gas equation (PV = nRT), it is evident that volume of a fixed mass of gas is constant if its pressure and temperature are held constant.

Since this question requires the gas to expand, then pressure and temperature cannot be constant. Hence, the assumption for constant pressure is invalid

Looking at what you have explained here:

V= nRT / P = constant at constant temp and pressure

therefore wouldnt it be possible for the temp or the pressure to be constant, I mean if volume of a fixed mass of ideal gas changes then isnt it possible for just the temp to change and the pressure remaining constant? (or vice versa?)

another quick q, if a system is isothermal does that mean that dT= 0?
« Last Edit: February 19, 2006, 11:59:25 AM by madscientist »
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#### Donaldson Tan

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« Reply #7 on: February 19, 2006, 04:39:05 PM »
Cp.dT - V.dp = 0 => Cp.dT = V.dp

For reversible adiabatic process involving a perfect gas,
Unless dT = 0, then dP = 0. This means if the reversible adiabatic process is isobaric, then it must be isothermal. However, from the perfect gas equation (PV = nRT), it is evident that volume of a fixed mass of gas is constant if its pressure and temperature are held constant.

Since this question requires the gas to expand, then pressure and temperature cannot be constant. Hence, the assumption for constant pressure is invalid

when you differentiate a constant, you get zero.

if T is constant, then dT = 0. if P is constant, then dP = 0.

Let's examine the final equation of my proof. For an isoentropic process, Cp.dT = V.dp

Since Cp and V are non-zero, then if dP = 0 then dT = 0, ie. temperature is constant, thus your assumption for constant pressure implies constant temperature. However, the volume of a fixed mass of perfect gas (V = nRT/P) will not change if its pressure and temperature are constant.

I mean if volume of a fixed mass of ideal gas changes then isnt it possible for just the temp to change and the pressure remaining constant? (or vice versa?)

Although it is possible for an ideal gas to expand under constant pressure, it is not possible for an ideal gas to expand isoentropically (dS=0) under constant pressure.
« Last Edit: February 19, 2006, 04:47:23 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006