Question

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.

Answer 1

Answer : The final temperature of gas is 266.12 K

Explanation :

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:

`\mu_(J,T)=((dT)/(dP))_H`

or,

`\mu_(J,T)=((dT)/(dP))_H\approx (\Delta T)/(\Delta P)`

As per question the formula will be:

`\mu_(J,T)=(T_2-T_1)/(P_2-P_1)` .........(1)

where,

`\mu_(J,T)` = Joule-Thomson coefficient of the gas =
`0.13K/atm`

`T_1` = initial temperature =
`19.0^oC=273+19.0=292.0K`

`T_2` = final temperature = ?

`P_1` = initial pressure = 200.0 atm

`P_2` = final pressure = 0.95 atm

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

`0.13K/atm=(T_2-292.0K)/((0.95-200.0)atm)`

`T_2=266.12K`

Therefore, the final temperature of gas is 266.12 K