Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressure of 0.95 atm. The Joule-Thomson coefficient of the gas is 0.13 K atm-1.

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

[tex]\mu_{J,T}=(\frac{dT}{dP})_H[/tex]

or,

[tex]\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}[/tex]

As per question the formula will be:

[tex]\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}[/tex] .........(1)

where,

[tex]\mu_{J,T}[/tex] = Joule-Thomson coefficient of the gas = [tex]0.13K/atm[/tex]

[tex]T_1[/tex] = initial temperature = [tex]19.0^oC=273+19.0=292.0K[/tex]

[tex]T_2[/tex] = final temperature = ?

[tex]P_1[/tex] = initial pressure = 200.0 atm

[tex]P_2[/tex] = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

[tex]0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}[/tex]

[tex]T_2=266.12K[/tex]

Therefore, the final temperature of gas is 266.12 K