Question

Consider the following statements:
(1) Joule-Thomson experiment is isenthalpic as well as adiabatic.
(2) A negative value of $\text{ }{{\mu }_{\text{JT}}}\text{ }$ (Joule Thomson coefficient corresponds to the warming of gas on expansion.
(3) The temperature at which neither cooling or heating effect is observed is known as inversion temperature.
Which of the above statements are correct (THIS QUESTION HAS MULTIPLE CORRECT OPTIONS)
A) 1 and 2
B) 1 and 3
C) 2 and 3
D) 1, 2 and 3

Answer 1

suppose a gas stored at a high-pressure allows it to expand to the region of the lower pressure. The expansion of gas through the plug appreciably cools down the gas. The sign of the JT coefficient $\text{ }{{\mu }_{\text{JT}}}\text{ }$ is used to determine whether gas undergoes the heating on expansion or not. The negative value suggests that gas expands at the expense of some work. From the JT experiment, the temperature below which a gas becomes cooler on expansion is known as the inversion temperature.

Complete step by step answer:
Consider a gas that is stored at high pressure is allowed to pass through a plug into a region of the low-pressure region, under an adiabatic condition, it gets cooled. The cooling effect due to the decrease in the kinetic energy of the gaseous molecule since a part of this energy is used up in overcoming the van der Waals force of attraction which is existing between the molecules during the expansion. This cooling effect of the gas is also known as the Joule Thomson effect. The expansion of gas occurs from the higher pressure to the lower pressure in such a way that in a compartment of higher pressure work is done on the system however in the lower [pressure region the work is done by the system. Since the expansion of gas is an adiabatic process, that system is not in a position to absorb or lose heat to the surrounding. thus joule –Thomson expansion of real gases occurs with constant internal energy but with constant enthalpy. Thus, the joule Thomson experiment is an iso enthalpy as well as an adiabatic process. The Joule Thomson coefficient $\text{ }{{\mu }_{\text{JT}}}\text{ }$ is equal to the change in the temperature with respect to change in pressure at constant enthalpy.it is represented as follows,$\text{ }{{\mu }_{\text{JT}}}\text{ }=\text{ }{{\left( \dfrac{\partial T}{\partial P} \right)}_{H}}\text{ }$ .
Suppose a gas expands in a vacuum thus doing no external work. However, some work will be done in separating the gas molecules against the forces of cohesion which exist between the molecules of the real gas. In these gases the temperature remains constant. During the isothermal expansion of a real gas there is no increase in the energy of the gas. Hence the joule Thomson coefficient is negative. Here, the real gas expands and warming of gas is observed on expansion. The temperature at which the joule-Thomson coefficient changes sign or the temperature at which gas does not undergo the following nor heating is known as the inversion temperature. The inversion temperature is given as follows,
$\text{ }{{\text{T}}_{\text{i}}}\text{ = }\dfrac{\text{2a}}{\text{Rb}}\text{ }$
Where $\text{ }{{\text{T}}_{\text{i}}}\text{ }$ is an inversion temperature, and b are van der Waals constants. Thus, all of the given statements are correct.

Hence, (A), (B), (C) and (D) all are correct options.

Note: The cooling effect or the JT effect is zero for ideal gas and nonzero for a real gas. When ideal gas expands in the vacuum there is neither absorption nor evolution of heat i.e. $\text{ q = 0 }$ .On the expansion of ideal gas, no work is done thus ideal gas does not expand. Similarly internal enthalpy of the expansion of ideal gas is equal to zero. Thus the ideal does not show the JT effect.