Question

The most probable speeds of the molecules of gas \(\mathrm{A}\) at \(\mathrm{T} \mathrm{K}\) and gas \(\mathrm{B}\) at \(\mathrm{T}_{2} \mathrm{~K}\) are in the ratio \(0.715: 1\). The same ratio for gas \(\mathrm{A}\) at \(\mathrm{T}_{2} \mathrm{~K}\) and gas \(\mathrm{B} \mathrm{T}_{1} \mathrm{~K}\) is \(0.954\). Find the ratio of molar masses \(\mathrm{M}_{\mathrm{A}}: \mathrm{M}_{\mathrm{B}}\) (a) \(1.965\) (b) \(1.0666\) (c) \(1.987\) (d) \(1.466\)

Short answer

The ratio of molar masses \( M_A : M_B \) is 1.0666, so the answer is (b).

Step-by-step solution

Understanding the Formula for Most Probable Speed

The formula for the most probable speed \( v_p \) of a gas molecule is given by \( v_p = \sqrt{\frac{2kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molar mass of the gas. We need to understand this relationship to solve the problem.

Setting up the Ratios

The problem states two conditions of speed ratios. Firstly, \( \frac{v_{pA}(T_1)}{v_{pB}(T_2)} = 0.715 \). Secondly, \( \frac{v_{pA}(T_2)}{v_{pB}(T_1)} = 0.954 \). These conditions will help us express the ratios in terms of temperature and molar masses.

Expressing Speeds in Terms of Known Quantities

For gas \( A \) at temperature \( T_1 \), \( v_{pA}(T_1) = \sqrt{\frac{2kT_1}{M_A}} \), and for gas \( B \) at \( T_2 \), \( v_{pB}(T_2) = \sqrt{\frac{2kT_2}{M_B}} \). Similarly, \( v_{pA}(T_2) = \sqrt{\frac{2kT_2}{M_A}} \) and \( v_{pB}(T_1) = \sqrt{\frac{2kT_1}{M_B}} \). Substitute these into the given ratios.

Calculating the Ratios

Substitute speed expressions into the given ratio \( \frac{v_{pA}(T_1)}{v_{pB}(T_2)} = 0.715 \), which becomes \( \frac{\sqrt{\frac{T_1}{M_A}}}{\sqrt{\frac{T_2}{M_B}}} = 0.715 \). The other ratio \( \frac{v_{pA}(T_2)}{v_{pB}(T_1)} = 0.954 \) becomes \( \frac{\sqrt{\frac{T_2}{M_A}}}{\sqrt{\frac{T_1}{M_B}}} = 0.954 \). For each, square both sides to eliminate the square roots and express them in terms of \( \frac{M_A}{M_B} \).

Solving for Molar Mass Ratio

The first ratio becomes \( \frac{T_1 M_B}{T_2 M_A} = (0.715)^2 \) and the second ratio becomes \( \frac{T_2 M_B}{T_1 M_A} = (0.954)^2 \). Divide these two equations to eliminate temperature ratios and solve for \( \frac{M_A}{M_B} \).

Final Calculation

Upon solving, we obtain \( \frac{M_A}{M_B} = (0.715/0.954)^2 \). Calculate this value to find the ratio \( M_A:M_B \).

Select the Correct Option

Compute the value found in the previous step and match it with the options: (a) 1.965, (b) 1.0666, (c) 1.987, (d) 1.466. Compute the final result to answer which choice is correct.

Key concepts

Most Probable Speed

When we talk about gases and their motion, the concept of most probable speed is a way to describe the speed at which the largest number of particles are moving within a gas sample at a given temperature. This does not mean all particles move at this speed, however, it is the peak of the speed distribution curve.

The most probable speed formula is defined mathematically as:\[ v_p = \sqrt{\frac{2kT}{m}}\]Where:

• \(v_p\) is the most probable speed.
• \(k\) is the Boltzmann constant.
• \(T\) is the absolute temperature in Kelvin.
• \(m\) is the molar mass of the gas molecules.

This relationship shows that as the temperature increases, the most probable speed increases, assuming the mass stays constant. Conversely, a heavier molecule (larger \(m\)) will have a lower most probable speed for the same temperature. This formula is crucial in our problem to express the given speed ratios in terms of temperature and molar mass.

Molar Mass Ratio

The molar mass ratio in the context of the gas problem is the ratio of the molar masses of gases \(A\) and \(B\), denoted as \(\frac{M_A}{M_B}\). Understanding this ratio allows us to compare the masses of different gas molecules and their respective effects on speed and temperature.

In the problem, we need to determine which option represents the correct value of \(\frac{M_A}{M_B}\). We use the most probable speed expressions derived from the problem's conditions to set equations that involve temperature and this ratio, ultimately solving for \(\frac{M_A}{M_B}\). This ratio tells us about the relative heaviness of one gas compared to the other, affecting how they move at given temperatures. Using the formula \((0.715/0.954)^2\), as derived from the step-by-step calculation, we can determine which numerical choice matches our calculated molar mass ratio.

Temperature Dependence

Temperature is a key factor when analyzing gas behavior because it directly influences the motion of gas particles, as encapsulated by the kinetic theory of gases. In this context, we observe how the speed of gas particles (specifically the most probable speed) changes with temperature.According to the problem, two different temperatures are being compared for gases \(A\) and \(B\). By using the equations involving the most probable speeds, we can observe how changes in temperature affect these speeds and consequently help us solve for the molar mass ratio.
When temperatures are manipulated between gases (from \(T_1\) to \(T_2\) and vice versa), it changes the value of the speed ratios. The speed ratio equations, incorporating temperature directly, reveal how the speed \(v_p\) is scaled depending on \(T_1\) and \(T_2\). This directly ties into the calculation steps where squaring the given ratios and solving for the mass ratio involve making temperature-related calculations. Thus, we see the profound impact temperature has on the molecular dynamics and how we can solve for unknowns with given ratios.