$$C_{p} - C_{v} = \frac{R}{M}$$ and $$C_{v}$$ are specific heats at constant pressure and constant volume respectively. It is observed that, $$C_{p} - C_{v} = a$$ for hydrogen gas and $$C_{p} - C_{v} = b$$ for nitrogen gas. The correct relation between $$a$$ and $$b$$ is:

For an ideal gas, the relation between its mass-specific heats is first written explicitly:

$$C_{p}-C_{v}=\frac{R}{M}$$

Here $$R$$ is the universal gas constant and $$M$$ is the molar mass of the gas expressed in $$\text{kg mol}^{-1}$$. Because $$R$$ is the same for every gas, the difference $$C_{p}-C_{v}$$ varies only with the reciprocal of the molar mass.

We are told that for hydrogen gas the difference is $$a$$, so

$$a=C_{p}-C_{v}\;(\text{for hydrogen})=\frac{R}{M_{\,\mathrm{H_2}}}.$$

For nitrogen gas the difference is $$b$$, hence

$$b=C_{p}-C_{v}\;(\text{for nitrogen})=\frac{R}{M_{\,\mathrm{N_2}}}.$$

To relate $$a$$ and $$b$$ we divide the first expression by the second:

$$\frac{a}{b}=\frac{\dfrac{R}{M_{\,\mathrm{H_2}}}}{\dfrac{R}{M_{\,\mathrm{N_2}}}} =\frac{M_{\,\mathrm{N_2}}}{M_{\,\mathrm{H_2}}}.$$

Now we substitute the molar masses in kilogram per mole:

$$M_{\,\mathrm{H_2}} = 2\;\text{g mol}^{-1}=2\times 10^{-3}\;\text{kg mol}^{-1},$$

$$M_{\,\mathrm{N_2}} = 28\;\text{g mol}^{-1}=28\times 10^{-3}\;\text{kg mol}^{-1}.$$

Putting these values into the ratio gives

$$\frac{a}{b}=\frac{28\times 10^{-3}}{2\times 10^{-3}}=\frac{28}{2}=14.$$

Therefore we finally obtain

$$a = 14\,b.$$

Hence, the correct answer is Option D.