Answer:
Joule-Thomson coefficient for an ideal gas:

Explanation:
Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.
Thus,
![\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H](https://tex.z-dn.net/?f=%5Cmu_%7BJ.T%7D%20%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20P%7D%20%5Cright%20%5D_H)
Also,


![dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT](https://tex.z-dn.net/?f=dH%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20T%7D%5Cright%20%5D_P%20dT%20%2B%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20dT)
Also,
is defined as:
![C_p = \left [\frac{\partial H}{\partial T}\right ]_P](https://tex.z-dn.net/?f=C_p%20%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20T%7D%5Cright%20%5D_P)

![dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT](https://tex.z-dn.net/?f=dH%3D%20C_p%20dT%20%2B%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20dT)
Acoording to defination, the ethalpy is constant which means 

![\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20%3D%20-C_p%5Ctimes%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_H)

![\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H](https://tex.z-dn.net/?f=%5Cmu_%7BJ.T%7D%20%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_H)

![\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20%3D-%5Cmu_%7BJ.T%7D%5Ctimes%20C_p)
For an ideal gas,
![\left [\frac{\partial H}{\partial P}\right ]_T = 0](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20%3D%200)
So,

Thus,
≠0. So,

Answer:
Tetrasulfur Dinitride S4N2
Tetrasulfur pentoxide S4N4
Tetrasulfur tetranitride N4S4
Tetrasulfur pentoxide S4N4
The answer is Helium. hope its helpful (:
Answer: D
Explanation: Bacteria thrives in warm and moist environments.