Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: karol.s on March 27, 2016, 05:43:31 PM
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Consider air expansion inside a cylinder. Assume that the volume and initial pressure is 1 ft^3 and 1500 PSI ABS respectively. If the expansion process is reversible and the path is given by P.V^1.4=constant. Calculate the total work done by the gas to reach the final volume of 8 ft^3, express the result in BTU if used:
a) The Van der Waals equation: (P + a.n ^ 2 / v ^ 2) * (v - nb) = nRT
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what I have is ...
P1* V1 ^1.4 = P2* V2 ^1.4
p2 = p 1 * (1/8)^1.4
=81.614 PSI
but i don`t know how to find the work with van der waals in an adiabatic process
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Again an inconsistent question... Because PV1.4=const needs a perfect gas that doesn't obey Van der Waal's equation, or at least, it imposes a=b=0. So one has to try to guess what you're supposed to do.
Do you know an expression for dW, the work during a small change of state, as a function of P and V?
Can you inject PV1.4=const in this, to express dW from P or V only?
Can you integrate dW as a function of P or V?
Where should you need Van der Waals?
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yes it was my mistake, I have to use T(v − b)^ R/cV = constant
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Well, once you've picked the proper formula (I doubt about the most recent one too), compute a dW as a function of P or V or T only and integrate it.
Does the question state a=0 somewhere?
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No, it doesn't says anything
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If the equation was
T×(V-b)R/Cv=const
this sounds like an adiabatic expansion with a=0.
That is, for the adiabatic expansion of a perfect gas you'd have
TCv/R×V=const where Cv is the heat capacity at constant volume
Now, "b" in Van der Waal's equation represents an added volume. You can represent it as the volume taken by the molecules' electrons. It resembles (but differs slightly) the volume of the liquid substance. This "b" isn't usable for the free movement of the molecules. This distinguishes them from a perfect gas whose molecules are points. Consequently, "b" is subtracted from the gas' volume in VdW equation.
"a" represents the attraction (usually) between the molecules.
If "a" is neglected, then the imperfect gas behaves like a perfect one where you just add a chunk of iron "b" in the volume. Simply replacing V by V-b everywhere, the gas behaves as a perfect one whose molecules have no volume. That's why you get
T×(V-b)R/Cv=const
I strongly suppose that a≠0 would not give that equation (but I won't check it, too long). Hence my question whether "a" is explicitly neglected. By the way, this would be reasonable in many technological situations like a hydraulic accumulator, where nitrogen's temperature exceeds clearly the critical point.
Now, supposing that a=0 if this is necessary, can you calculate the work? Same adiabatic expansion as a perfect gas, but with a chunk of iron "b" in the volume.
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By the way, these two are your allies for the adiabatic expansion of a perfect gas:
TCv/R×V=const
T-Cp/R×P=const
because
- they're easy to memorize: CV goes with V and CP with P.
- CV/R represents half the degrees of freedom of a perfect gas
(more accurately: each translation and rotation brings R/2 to CV, each excited vibration adds R) - CP/R is 1+CV/R, since H=U+RT
- The ratio of both equations gives you PVCp/Cv=const where you recognize γ.
- It gives you quickly the temperature from P or V ratios, and the temperature tells you easily the work, the heat - in the case of a perfect gas, pretty much everything.
The previous 1.4 stands for a diatomic gas like nitrogen, oxygen and by extension air. 1.40 is even nicely accurate under usual conditions, useful to have in mind. According to the list:
- 3 transations bring 3/2×R to CV
- 2 rotations (no third rotation since N2 and O2 are linear) bring 2/2×R
- N2 and O2 don't vibrate at room temperature nor moderate heat, so nothing to add
- Hence CV=2.5×R, CP=3.5×R and γ=3.5/2.5=1.40