Question

A real gas undergoes throttling from 50 bar to 10 bar from an initial temperature of 300 K. If \(\mu_{JT} = -0.05 \text{ K/bar}\) for the initial condition, what is the approximate value of the exit temperature?

• A. \(298 \text{ K} \)
• B. \(302 \text{ K} \)
• C. \(300 \text{ K} \)
• D. \(304 \text{ K} \)

Hint:

When $\mu_{JT}$ is negative, the gas warms up during expansion ($\Delta P$ is always negative in a flow restriction). A quick qualitative check reveals that since $\mu_{JT} \lt 0$, $T_2$ must be greater than $T_1$. This instantly eliminates options (A) and (C).

Answer 1

The Correct Option is B

Step 1: Note that this is a real gas, not an ideal one.
For an ideal gas, throttling would leave the temperature unchanged, but here we are told \(\mu_{JT} = -0.05 \text{ K/bar}\), a negative value. A negative Joule-Thomson coefficient means this gas, at this state, is on the side of its inversion curve where dropping pressure actually raises temperature instead of cooling it.

Step 2: Find the pressure drop the gas experiences.
\[ \Delta P = P_2 - P_1 = 10 - 50 = -40 \text{ bar} \]

Step 3: Apply the coefficient to get the temperature change.
\[ \Delta T = \mu_{JT} \times \Delta P = (-0.05) \times (-40) = 2 \text{ K} \] Since both the coefficient and the pressure change are negative, their product is positive, confirming the gas heats up rather than cools. Adding this rise to the initial temperature: \[ T_2 = 300 + 2 = 302 \text{ K} \] \[ \boxed{T_2 \approx 302 \text{ K}} \] This matches option 2.