Question

Experiments over a limited range of pressure and temperature not far from normal conditions show that the Joule-Thomson coefficient for air may be represented by this empirical equation:

μ = - 0.1975 + 138/T - 319p/T^2 K atm-1 (a) Evaluate μ and du/dT at constant P for air at 60 ̊C and 1 atm.

(b) Will air in a tank at 5 atm and 25 ̊C become cooler or warmer when the tank valve is

opened and the air expands into a room at 25 ̊C and 1 atm. Explain.
(c) Estimate the approximate temperature at which air would have to be for its temperature to

change in the opposite way when its pressure was changed from 5 atm to 1 atm.

Note: Since the empirical relation given here is valid only over a limited range, precision in this estimate is not meaningful; you can get an adequate first approximation by ignoring the third term in the equation.)

3(a). 0.214 K atm-1; -0.00123 atm-1 (b) cooler; (c) about 700 K ( ≈ 430 ̊C)

Answer 1

The equation for Joule-Thomson coefficient for air is given as

μ = - 0.1975 + 138/T - 319p/T² K atm^-1

a. T = 60 ^oC, i.e. 60 + 273 = 333 K

and p = 1 atm

Now, μ = - 0.1975 + 138/333 - 319*1/333² K atm^-1

i.e. μ = 0.214 K atm^-1

And dμ/dT = 138*(-1/T²) - 319p*(-2/T³) atm^-1

= 138*(-1/333²) - 319*1*(-2/333³) atm^-1

i.e. dμ/dt = -0.00123 atm^-1

b. T = 25 ^oC = 25+273 = 298 K and p₁ = 5 atm

Now, μ₁ = - 0.1975 + 138/298 - 319*5/298² K atm^-1

i.e. μ₁ = 0.248 K atm^-1

And T = 25 ^oC = 25+273 = 298 K and p₂ = 1 atm

Now, μ₂ = 0.214 K atm^-1

Therefore, dμ = μ₂ - μ₁ = 0.214 - 0.248 = -0.034

Here, dμ is a negative value, hence cooling takes place.

c. {- 0.1975 + 138/700 - 319*1/700²} - {- 0.1975 + 138/700 - 319*5/700²}

= -0.001 -(-0.004)

= 0.003

Here, dμ is a positive value, hence heating takes place, which is the change in opposite way.

Hence, T = 700 K is verified as the correct answer