Joule Thomson effect: A typical question
Question:
“I'm not sure about the Joule Thomson effect on gas mixtures. I looked on the internet but couldn't find the concept. The thing is, when I flash ammonia, it cools to -33 degrees Celsius and nitrogen to -10 degrees Celsius. However, when I flash a 1:1 mole ratio of nitrogen and ammonia, it cools to -52degC. I couldn't figure out what was going on. Please advise on how to approach this problem."
I am assuming that it is not a case for Flash evaporation rather it is a case of the Joule Thomson effect. Flash evaporation occurs when a saturated liquid stream undergoes a reduction in pressure by passing through a throttling valve or other throttling devices. There is no liquid in this case.
Joule Thomson effect
Briefly, in thermodynamics, the JT effect describes the temperature change of a real gas or liquid when it is forced through a valve or porous plug while keeping it insulated so that no heat is exchanged with the environment. This procedure is called a throttling process or JT process. At room temperature, all gases except hydrogen, helium, and neon cool upon expansion by the Joule–Thomson process. JT is an adiabatic process. It acts in iso-enthalpic manner. This cooling of the gas is basically due to the decrease in the kinetic energy of the gaseous molecules as some part of this kinetic energy is utilized in overcoming the intermolecular van der Waals force of attraction during expansion. Kinetic energy converts to potential energy keeping the enthalpy constant which cools the gas.
Mechanism of JT effect
JT effect is approximated as an Isenthalpic.
How is JT isenthalpic?
The gas cools at constant enthalpy. Any temperature change is the function of internal energy – kinetic and potential energy. Cooling observed in JT suggests that there is a change in the internal energy in JT at constant enthalpy
At two points 1 and 2
The work is done by the system at the expense of
internal energy only. Let the internal energy of the system changes from U1 to U2
where U is internal energy, P is pressure, and V is volume. the change in PV represents the work done
The net work is done when a mass m of the gas expands
W = mV1P1 - m V2P2
as the gas expands from P1VI to P2V2, it does a W amount of work.
From the first law of thermodynamics dU-dW = dQ
In the JT process, dQ=0 as per the definition. JT expansion is an adiabatic process
So, dU = dW
Since dH = dU +PV
Therefore [mU2 -mU1] - [mV1P1 - m V2P2] =0
This gives mH1 = mH2,
H1=H2 meaning Isenthalpic process.
LinkedIn question:
“I'm not sure about the Joule Thomson effect on gas mixtures. I looked on the internet but couldn't find the concept. The thing is, when I flash ammonia, it cools to -33 degrees Celsius and nitrogen to -10 degrees Celsius. However, when I flash a 1:1 mole ratio of nitrogen and ammonia, it cools to -52degC. I couldn't figure out what was going on. Please advise on how to approach this problem."
Explanation
A brief explanation of intermolecular forces between molecules
Very broadly we see the following types of intermolecular forces in molecules
London forces [ temporary dipoles]: It applies to every type of molecule including non-polar molecules
The London dispersion force is the weakest intermolecular force. The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction. Because of the constant motion of the electrons, an atom or molecule can develop a temporary (instantaneous) dipole when its electrons are distributed unsymmetrically about the nucleus.
Van der Walls force: It applies to all real gas molecules
Van der Waals forces include attraction and repulsions between atoms, molecules, and surfaces, as well as other intermolecular forces. The force results from a transient shift in electron density. Specifically, the electron density may temporarily shift more greatly to one side of the nucleus. This shift generates a transient charge that a nearby atom can be attracted to or repelled by.
Dipole-Dipole intermolecular forces [permanent dipoles]
Dipole-dipole forces are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule. This primarily arises when one atom in the molecule is more electronegative than another. In simple sense H- bonds are one such special type of bond
In the question, there are two gases, NH3 and N2
Ammonia gas
In NH3 the nitrogen atom is more electronegative [3.04] than the hydrogen atom, [2.2]. ?Since nitrogen is more electronegative than hydrogen, it pulls the free electron pair more strongly towards itself, acquiring a partial negative charge. Thus, besides other types of intermolecular forces, NH3 has dipole-dipole forces.
Nitrogen gas
In N2, one N atom cancels the polar effect of another N. So, in N2 there is no charge polarity. However, N2 has other types of intermolecular attractions as explained.
Ammonia is a more polar molecule than N2. Ammonia has more intermolecular forces
Proof
Boiling point of NH3 = -33.34 degc
Boiling point of N2 = - 195.8 degc
The higher boiling point of NH3 establishes that it has more intermolecular forces than N2
What do larger intermolecular forces of NH3 mean?
Why NH3 shows more cooling than N2 [assumed under the same conditions]?
Larger intermolecular forces in an adiabatic process suggest more internal energy consumption. Therefore, NH3 cools more than N2.
Next question regarding mixing NH3 and N2 mixing in [1:1] mole ratio
When N2 is mixed with NH3, you introduce a new NH3-N2 intermolecular attraction. That would consume more internal energy than N2 and NH3 would consume when expanded. This explains why the cooling rose to -52 degc.
Proof
Let us take the example of air that contains two gases, 79% N2 and 21% O2 [ traces of other gases ignored]
Liquefaction temperature of N2 = -195.8 degc
Liquefaction temperature of O2 = -180 degc [ O2 has higher liquefaction temperature because it is bigger in size than N2, 32 vs 28] and therefore it has more intermolecular forces]
Liquefaction temperature of air = - 196 degc.
Weighted average = 0.79 x [-195.8] + 0.21 x [ - 180] = [ - 154.7 -37.8] = - 192.5
Actual liquefaction temperature of air = -196 deg
That clearly shows that when two molecules with different polarities are mixed the relation between intermolecular forces no longer remains linear.