Nitrogen gas is at $$300\,^\circ\text{C}$$ temperature. The temperature (in K) at which the rms speed of a $$\text{H}_2$$ molecule would be equal to the rms speed of a nitrogen molecule, is........... (Molar mass of $$\text{N}_2$$ gas 28 g).

First, we convert the given Celsius temperature of nitrogen gas into the Kelvin scale, because all gas-kinetic formulae are written for absolute temperature. By definition,

$$T\;(\text{in K}) \;=\; T\;({}^\circ\text{C}) + 273.$$

Here the nitrogen temperature is $$300\,{}^\circ\text{C}$$, so

$$T_{\text{N}_2}=300+273=573\;\text{K}.$$

For any ideal gas the root-mean-square (rms) speed is given by the well-known kinetic-theory relation

$$v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}},$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass (expressed in the same mass unit for every gas in the comparison).

We are told that the rms speed of a hydrogen molecule $$(\text{H}_2)$$ must equal the rms speed of a nitrogen molecule $$(\text{N}_2)$$. Therefore we set the two expressions equal:

$$\sqrt{\dfrac{3R\,T_{\text{N}_2}}{M_{\text{N}_2}}}\;=\;\sqrt{\dfrac{3R\,T_{\text{H}_2}}{M_{\text{H}_2}}}.$$

First we square both sides to remove the square roots, obtaining

$$\dfrac{3R\,T_{\text{N}_2}}{M_{\text{N}_2}}\;=\;\dfrac{3R\,T_{\text{H}_2}}{M_{\text{H}_2}}.$$

The common factor $$3R$$ cancels out, leaving the simple proportionality

$$\dfrac{T_{\text{N}_2}}{M_{\text{N}_2}}\;=\;\dfrac{T_{\text{H}_2}}{M_{\text{H}_2}}.$$

We now solve for the unknown temperature $$T_{\text{H}_2}$$ of hydrogen:

$$T_{\text{H}_2} = T_{\text{N}_2}\,\dfrac{M_{\text{H}_2}}{M_{\text{N}_2}}.$$

Substituting the numeric values, we use the molar mass of nitrogen as $$M_{\text{N}_2}=28\;\text{g mol}^{-1}$$ (given) and the customary molar mass of hydrogen as $$M_{\text{H}_2}=2\;\text{g mol}^{-1}$$. Hence,

$$T_{\text{H}_2} = 573\;\text{K}\,\times\,\dfrac{2}{28}.$$

Now we simplify the fraction:

$$\dfrac{2}{28} = \dfrac{1}{14}.$$

So,

$$T_{\text{H}_2} = 573\;\text{K}\,\times\,\dfrac{1}{14}.$$

Carrying out the division,

$$T_{\text{H}_2} = \dfrac{573}{14}\;\text{K} = 40.93\;\text{K}.$$

Rounding to the nearest whole number in Kelvin, we have $$T_{\text{H}_2}\approx 41\;\text{K}.$$

Hence, the correct answer is Option 41.