A real gas expands against zero pressure there, observe no change in enthalpy of the gas but cooling is observed, this effect is not observed when real gas is replaced by ideal gas. justify this statement with your point of view. show that cooling effect in the case of real gas is isenthalpic process. also, prove that in the case of ideal gas (t/p)h = 0
Answers
r∝
M
1
∴
r
O
2
r
H
2
=
M
H
2
M
O
2
=
1
16
=4
∴r
H
2
=4×r
O
2
Joule-Thomson coefficients of quantum ideal-gases. The temperature drop of agas divided by its pressure drop under constant enthalpy conditions is called theJoule-Thomson coefficient (JTC) of the gas. The JTC of an ideal gas is equal to zero since its enthalpy depends on only temperature.
[
∂P
∂T
]
H
=0 (for ideal gas)
Because T∝P (for ideal gas)
∴
P
T
=K (constant)
Answer:
See the explanation below.
Explanation:
When a non-ideal gas suddenly expands from higher pressure to low pressure, there is a temperature change.
This is called Joule-Thomson Effect. It is an adiabatic effect. The temperature of a real gas is either decreased or increased by letting the gas expand freely at constant enthalpy, the temperature may either increase or decrease, depending on the initial pressure and temperature. For any given pressure, a real gas has an inversion temperature above which the expansion at constant enthalpy causes the temperature to rise and below which it causes cooling. For an ideal gas, there are no intermolecular forces, so no temperature change is expected when the distance between the molecules changes.