Question

The ratio of universal gas constant and molar mass of gas is called molar gas constant. The value of molar gas constant is greater for?
A.He
B. $N_2$
C. $H_2$
D.Same for all

Answer 1

Hint: The gas constant R is 8.314 J/mol.k. We know the molar mass of He, $N_2$ and $H_2$. Therefore, to find the value of the molar mass gas constant of the gas, we will calculate $${\displaystyle \frac{R}{M}}$$.

Complete step by step answer:
The ratio of molar gas constant is denoted by (R) to the molar mass denoted by M of the gas mixture is called the molar gas constant. Denoted by $R_{specific}$ Mathematically expressed as:
 $R_{specific}={\displaystyle \frac{R}{M}}$
The gas constant R is 8.314J/mol.k.
Molar gas constant = universal gas constant​ / molar mass of gas
So, molar gas constant is directly proportional to molar mass of the gas.

So, molar gas constant $$\propto$$ molar mass1

More molar mass less is molar gas constants
molar mass $$He=4$$
molar mass $N_2=28$
molar mass $H_2=2$
The value of molar gas constant of $$He={\displaystyle \frac{R}{M}}={\displaystyle \frac{8.314}{4}}=2.0785$$

The value of molar gas constant of $N_2={\displaystyle \frac{R}{M}}={\displaystyle \frac{8.314}{28}}=0.2969$

The value of molar gas constant of $H_2={\displaystyle \frac{R}{M}}={\displaystyle \frac{8.314}{2}}=4.157$

So, molar gas constant is more for $H_2$ as it has less molar mass.

Therefore, the correct answer is option (C).

Note: Molar gas constant which is denoted by symbol R is a fundamental physical constant arising in the formulation of the general gas law. For an ideal gas, the pressure(p) times the volume (V) of the gas divided by its absolute temperature (T) is a constant. When one of these three is changed for a given mass of gas then at least one of the other two undergoes a change so that the expression $${\displaystyle \frac{PV}{T}}$$ will be constant. The constant is the same for all gases, provided that the mass of gas being compared is one mole, or one molecular weight in grams. For one mole, it is therefore, $${\displaystyle \frac{PV}{T}}=R$$.