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**Hint:**Brush up the important concepts of thermodynamics. Recollect Graham’s law of diffusion. Remember, when work is done adiabatically, heat is not supplied on the system.

**Complete answer:**

-Let’s have a look at the statements given in the question one by one.

(A) Hydrogen gas, ${{H}_{2}}$ diffuses four times faster than oxygen gas, ${{O}_{2}}$.

Molecular weight of hydrogen gas, ${{H}_{2}}$ is 2g whereas molecular weight of oxygen, ${{O}_{2}}$ is 32g.

According to Graham’s law, “The rate of diffusion or movement of gaseous molecules is inversely proportional to the square root of its molecular weight.”

\[\text{Rate of diffusion }\propto \dfrac{1}{\sqrt{M}}\]

Square root of molecular weight of ${{H}_{2}}$ gas is $\sqrt{2}$

Square root of molecular weight of ${{O}_{2}}$ gas is $\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}$

\[\dfrac{\text{Rate of diffusion of }{{\text{H}}_{2}}}{\text{Rate of diffusion of }{{\text{O}}_{2}}}=\dfrac{4\sqrt{2}}{\sqrt{2}}=4\]

\[\text{Rate of diffusion of }{{\text{H}}_{2}}=4\times \text{Rate of diffusion of }{{\text{O}}_{2}}\]

Hence, it can be concluded that statement (A) is correct. Hydrogen gas, ${{H}_{2}}$ diffuses four times faster than oxygen gas, ${{O}_{2}}$.

(B) The temperature of a real gas changes when it expands adiabatically in vacuum.

This statement is correct. The temperature of a real gas does change when it expands adiabatically in vacuum because during adiabatic expansion no heat enters the system so energy is utilized from the gas itself and thus, there is a drop in temperature.

(C) An ideal gas undergoes a cooling effect when it suffers an adiabatic expansion in a vacuum.

When an ideal gas expands in vacuum adiabatically, no heat is supplied to the system externally. So, the system uses energy from the gas itself which leads to lowering of temperature and there is a cooling effect.

(D) The Joule-Thomson coefficient ${{\left[ \dfrac{dT}{dP} \right]}_{H}}$ of an ideal gas is zero.

The ratio of change in temperature of gas to change in its pressure is known as Joule-Thomson coefficient of the gas. Its enthalpy depends only on temperature.

${{\left[ \dfrac{dT}{dP} \right]}_{H}}=0$

Therefore, all the statements given in the question hold true.

**Note:**Don’t get confused that all the statements given in the statement are true. In an adiabatic expansion, heat is never supplied to the system and the system uses internal energy for getting the work done. In the Joule-Thomson coefficient, enthalpy depends on temperature and if enthalpy is constant then change in temperature with respect to pressure automatically becomes equal to zero.

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