When the temperature of a gas is raised from \(27^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the percentage increase in the rms velocity of the molecules will be (A) \(15 \%\) (B) \(17.5 \%\) (C) \(10 \%\) (D) \(20 \%\)

First, let's convert the initial temperature from Celsius to Kelvin since rms velocity is typically expressed in terms of Kelvin. We can use the formula: \[T(K) = T(^\circ C) + 273.15\] So, the initial temperature in Kelvin is: \[T_1(K) = 27^\circ C + 273.15 = 300.15 K\] Next, we will use the formula for the rms velocity, which is given by: \[v_\text{rms} = \sqrt{\frac{3kT}{m}}\] where \(v_\text{rms}\) is the rms velocity, \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of a single gas molecule. We don't know the mass of a single gas molecule, but since we only need the percentage increase in velocity, we can still proceed without knowing the value of mass. The initial rms velocity is: \[v_{rms 1} = \sqrt{\frac{3kT_1}{m}}\]

Now we convert the final temperature from Celsius to Kelvin: \[T_2(K) = 90^\circ C + 273.15 = 363.15 K\] Then, obtain the final rms velocity: \[v_{rms 2} = \sqrt{\frac{3kT_2}{m}}\]

Now that we have the initial rms velocity (\(v_{rms 1}\)) and the final rms velocity (\(v_{rms 2}\)), we can compute the percentage increase using the formula: Percentage increase = \(\frac{(v_{rms 2} - v_{rms 1})}{v_{rms 1}} × 100\%\) We can simplify the equation by substituting the expressions of \(v_{rms 1}\) and \(v_{rms 2}\) into the formula above: Percentage increase = \(\frac{\sqrt{\frac{3kT_2}{m}} - \sqrt{\frac{3kT_1}{m}}}{\sqrt{\frac{3kT_1}{m}}} × 100\%\) As we can see, the term \(\sqrt{\frac{3k}{m}}\) is common in all expressions, so we can simplify the expression by canceling out this term: Percentage increase = \(\frac{\sqrt{T_2} - \sqrt{T_1}}{\sqrt{T_1}} × 100\%\) Now, substitute the values of \(T_1\) and \(T_2\) into the equation: Percentage increase = \(\frac{\sqrt{363.15} - \sqrt{300.15}}{\sqrt{300.15}} × 100\% \approx 17.5\%\) Thus, the percentage increase in the rms velocity of the molecules is approximately 17.5%, so the correct answer is (B).