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Step-by-Step Solution
Step 1: Recall the relation between molar specific heats
For an ideal gas, the molar specific heats at constant pressure and volume satisfy the relation
$ C_p - C_v = R $.
Here, $C_p$ and $C_v$ refer to the molar specific heats (in terms of per mole of the gas), and $R$ is the universal gas constant.
Step 2: Extend the relation to n gram moles
When dealing with n gram moles of a gas, the corresponding difference in specific heats (per gram of the gas) becomes
$ \displaystyle \frac{R}{n} $.
This is because the total molar specific heat difference $R$ for 1 mole is now distributed over n moles.
Step 3: Apply to hydrogen (n = 2)
For hydrogen gas (molecular mass = 2 g/mol), the number of gram moles present in 1 g of hydrogen is
$ n = 2 $.
Thus,
$
C_p - C_v = \frac{R}{2} = a.
$
Step 4: Apply to nitrogen (n = 28)
For nitrogen gas (molecular mass = 28 g/mol), the number of gram moles present in 1 g of nitrogen is
$ n = 28 $.
Hence,
$
C_p - C_v = \frac{R}{28} = b.
$
Step 5: Relate a and b
We now compare the two expressions for $a$ and $b$:
$
a = \frac{R}{2}, \quad b = \frac{R}{28}.
$
So,
$
\frac{a}{b} = \frac{\frac{R}{2}}{\frac{R}{28}} = \frac{28}{2} = 14.
$
Thus,
$
a = 14\,b.
$
Final Answer
The correct relation between $a$ and $b$ is
$ a = 14\,b.
$
