We are given the ratio of the vapour densities as: \[ \frac{\rho_1}{\rho_2} = \frac{4}{25} \]
Step 2: Relate Vapour Density Ratio to r.m.s. Velocity RatioWe know that the ratio of r.m.s. velocities \( v_1 \) and \( v_2 \) is related to the ratio of vapour densities by the formula: \[ \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]
Step 3: Calculate the Ratio of r.m.s. VelocitiesSubstituting the given vapour density ratio: \[ \frac{v_1}{v_2} = \sqrt{\frac{25}{4}} = \frac{5}{2} \]
Final Answer: \[ \frac{v_1}{v_2} = \frac{5}{2} \]Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).