# the Joule Thomson coefficient

Essay by ash21College, Undergraduate November 2014

Keywords

# Objective:

In this experiment, we are aiming to determine the Joule Thomson coefficient ( ) of CO2 and N2 gases.

# Equipments:

Joule-Thomson apparatus, digital temperature meter, two temperature probes. Two vacuum rubbers tubing, reduce valves for CO2 and N2 two steel cylinders for CO2 and N2 gases, 10 liters capacity each.

# Theory:

In real gases, the internal energy U is composed of a thermo kinetic content : the potential of the intermolecular forces of attraction. This is negative and tends towards zero as the molecular distance increases. In real gases, the internal energy is therefore a function of the volume while the internal energy for ideal gas is a function of temperature only and hence:

During adiabatic expansion ( and during which also no external work done, the overall internal energy remains unchanged, with the result that potential energy increases at the expanses of the thermo-kinetic content and the gas cools.

H1=U1+P1V1=U2+P2+V2=H2

In this equation, PV is the work performed by an imaginary piston during the flow of a small amount of gas by a change in piston 1 to 2 or piston 3 to 4 (Fig-2).

In real gases, the displacement work PV does not equal to the displacement work equal the displacement work P V: in this case: P1V1 P2V2

# Procedure:

First of all , the reducing valves of the steel cylinder area screwed and the main valves tightness area checked too. The steel cylinders are put on a secure location. The vacuum between the reducing valve and the Joule-Thomson apparatus is attached with hose tube clips. A temperature probe on the un-pressurized side to inlet 2 of the temperature measurements apparatus.

# Results:

 P N2 CO2 1 0 0.52 0.9 -0.005 0.49 0.8 -0.015 0.46 0.7 -0.013 0.42 0.6 -0.015 0.39 0.5 -0.017 0.31 0.4 -0.024 0.24 0.3 -0.025 0.15 0.2 -0.026 0.08 0.1 -0.028 0.01 0 -0.032 0

# Graph:

 P (bar) 0 0 0.9 0.09 0.8 0.13 0.7 0.16 0.6 0.22 0.5 0.24 0.4 0.27 0.3 0.3 0.2 0.33 0.1 0.4 0 0.43

# Calculations:

From the slope equation in the graph we get:

Y= 7 * 106 -0.0614 for CO2 y= 7 * 10-7 X-0.2364 for N2

Y=mx +c y=mx+c

Slope = 7 * 10-6- Slope = 7 * 10-7

Therefore,

ÃÂµJT = 7 * 10-7 K\Pa for N2

ÃÂµJT = 7 * 10-6- K\Pa for CO2

Van Der Waals

-For Co2

= 9.43 * 10-4 K/Pa

-For N2

=-7.4 * 10-3 K/Pa

The literature value for CO2: is ÃÂµCO2= 1.16 * 10 -5 K/Pa

The literature valve for N2 : is ÃÂµN2=

Co2:

Experimental value VS. Literature valve

7*10 -6 K/Pa Vs 1.16 * 105 K/Pa

=

= 39.6 %

N2:

Experimental value Vs Literature value

7* 10-7 K/Pa Vs 1.16 * 10-5 K/Pa

= * 100

= 72 %

# Conclusion:-

From this experiment we can determine the Joule Thomson coefficient ( ÃÂµ) of CO2 and N2 gases. Also The experimenting room and the experimental apparatus

Must be in a thermal equilibrium at the start of the measurement.

The experimental apparatus should be kept out of direct

Sunlight and other sources of heating or cooling.