Thermodynamics - enthalpy problem

[simina11]

Hi, I have a practice problem in thermo that I have some difficulties with:

The PVT relation for helium is given by the following equation of state:

P=RT/V-b, where b=2.3*10^5 m^3/mol

Through an isenthalpic (H=0) throttling process, the gas is let to expand from a pressure of 10^8 Pa at 300 K to a pressure of 10^5 Pa. Calculate the final temperature of the gas. Cp can be considered constant at a value of 5/2R.

I started by using the eq. for enthalpy

dH=0=CpdT + (V-T(dV/dT))dP and finding (dV/dT) from the PVT relation given which resulting in R/P

Replacing R/P in the dH eq. we get:

0=CpdT + (V-T(R/P))dP

Wouldn't this cancel the dP term since TR/P =V?

Or should I not cancel them and just continue to integrate?

I'm a bit confused and could use some help.

Thank you in advance

[Enthalpy]

Hi simina11, welcome here!

 

Are any prentheses missing in P=RT/V-b ?

Does TR/P =V apply here despite b?

[studiot]

b has the units of volume so the formula per mole is

 

[math]P = \frac{{RT}}{{\left( {V - b} \right)}}[/math]

It is a cut down Van der Waal's equation.

 

 

Edited February 17, 2015 by studiot

[studiot]

Hello again simina, have you abandoned Science Forums?

 

I left the question for Enthalpy to complete with you, as is seemed particularly apt from the zero enthalpy condition.

 

But since no one else has added more here are my thoughts.

 

Firstly you need to be much more careful about careless errors, you have two in this question, this is a pity because you are on the right track.

 

 

Wouldn't this cancel the dP term since TR/P =V?

Or should I not cancel them and just continue to integrate?

 

Why would this be?

You need to check your maths more carefully, where did you get the equation from?

 

[math]\int\limits_{{h_1}}^{{h_2}} {dh} = \int\limits_{{T_1}}^{{T_2}} {{C_p}dT} + \int\limits_{{P_1}}^{{P_2}} {\left[ {V - T{{\left( {\frac{{\partial V}}{{\partial T}}} \right)}_P}} \right]} dP[/math]

Which you have correct.

 

But then you are too quick in your substitution from your equation of state.

Yes the R/P terms 'cancel' (though I don't like this word like this) but that does not destroy the equation when you integrate.

 

Thus

[math]V = \frac{{RT}}{P} + b[/math] and [math]{\left( {\frac{{\partial V}}{{\partial T}}} \right)_P} = \frac{R}{P}[/math]

Substituting

 

[math]\int\limits_{{h_1}}^{{h_2}} {dh} = \int\limits_{{T_1}}^{{T_2}} {{C_p}dT} + \int\limits_{{P_1}}^{{P_2}} {\left[ {\frac{{RT}}{P} + b - T\frac{R}{P}} \right]} dP[/math]

So all three intergral terms still exist, therefore performing the integration

[math]\int\limits_{{h_1}}^{{h_2}} {dh} = \int\limits_{{T_1}}^{{T_2}} {{C_p}dT} + \int\limits_{{P_1}}^{{P_2}} {\left[ { + b} \right]} dP[/math]

[math]\Delta h = 0 = {C_p}\Delta T + b\Delta P[/math]

 

rearranging

[math]\Delta T = - \frac{b}{{{C_p}}}\Delta P[/math]

 

Now care is needed at this point because deltaP is negative here therefore deltaT is positive, which is correct since the Helium is above its inversion temperature.

 

Someone has mistranscribed the value of b by a factor of 10¹⁰!

 

b should be 2.3 ^-5x10, not 2.3 x 10⁵ as you had it.

 

(don't you think the forum superscipt and subscript is better than the carat ^ symbol?)

When you put these values into the final equation you get a resonable answer for the temperature increase of the Helium.

Further care is needed as tio the value of R employed.

Do we mean the engineer's R (the universal gas constat divided by the molar mass) or the universal constant itself?

 

I will leave you to ponder this one.

Edited February 25, 2015 by studiot

[studiot]

I see that there have been quite a few views since I posted so here is some background.

 

Engineers use tables/charts of the

 

Joule Thompson coefficient, which is basically the ratio of delta T to delta P

 

Actually it is the slope of the constant enthalpy line on a T_P diagram.

 

 

http://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect

 

This reference has a value of 0.055 ^oK /bar for this coefficient at 300^oK

 

The question basically drops from 10⁸ to 10⁵ Pa ie (1000 - 1) x10⁵ Pa or 1000 bar,

 

Which implies a temperature rise of 55^oK

 

Here is someone reporting burning himself on the temperature rise from 200 bar cylinder of Helium throttling to 1bar.

 

https://www.physicsforums.com/threads/i-got-burned-by-a-helium-tank-valve-how.627972/

 

An typo crept into the previous post

 

b should be 2.3 x 10^-5

Edited February 28, 2015 by studiot