Experiments over a limited range of pressure and temperature not far from normal

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:

= - 0.1975 + 138/T - 319p/T^2 K atm-1 (a) Evaluate 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)

Explanation / Answer

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

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

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

and p = 1 atm

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

i.e. = 0.214 K atm-1

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

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

i.e. d/dt = -0.00123 atm-1

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

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

i.e. 1 = 0.248 K atm-1

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

Now, 2 = 0.214 K atm-1

Therefore, d = 2 - 1 = 0.214 - 0.248 = -0.034

Here, d 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, d 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

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