Question

G1 and G2 are two ideal gases at temperatures \(T_1\) and \(T_2\), respectively. The molecular weight of the constituents of G1 is half that of G2. If the average speeds of the molecules of both gases are equal, then assuming Maxwell-Boltzmann distributions for the molecular speeds, the ratio \(\frac{T_2}{T_1}\) is \rule{1cm{0.15mm}.}

Hint:

For any characteristic speed (rms, average, or most probable) of an ideal gas, the speed is always proportional to \(\sqrt{T/M}\). Therefore, if the speeds of two gases are equal, the ratio \(T/M\) must be the same for both. This provides a quick way to solve such problems: \(\frac{T_1}{M_1} = \frac{T_2}{M_2}\).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The problem relates the temperature and molecular weight of two different ideal gases using their average molecular speed. The average speed of gas molecules is determined by the gas's temperature and the mass of its constituent molecules, as described by the Maxwell-Boltzmann distribution.
Step 2: Key Formula or Approach:
The average speed \(\langle v \rangle\) of molecules in an ideal gas at temperature T is given by the formula derived from the Maxwell-Boltzmann distribution: \[ \langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}} = \sqrt{\frac{8 R T}{\pi M}} \] where \(k_B\) is the Boltzmann constant, \(m\) is the mass of a single molecule, \(R\) is the universal gas constant, and \(M\) is the molar mass (or molecular weight) of the gas.
Step 3: Detailed Explanation:
Let the properties of gas G1 be denoted by subscript 1 and gas G2 by subscript 2. We are given: Molecular weight relation: \(M_1 = \frac{1}{2} M_2\) Average speeds are equal: \(\langle v_1 \rangle = \langle v_2 \rangle\) Using the formula for average speed for each gas: For G1: \(\langle v_1 \rangle = \sqrt{\frac{8 R T_1}{\pi M_1}}\) For G2: \(\langle v_2 \rangle = \sqrt{\frac{8 R T_2}{\pi M_2}}\) Since \(\langle v_1 \rangle = \langle v_2 \rangle\), we can set their expressions equal to each other: \[ \sqrt{\frac{8 R T_1}{\pi M_1}} = \sqrt{\frac{8 R T_2}{\pi M_2}} \] Squaring both sides of the equation: \[ \frac{8 R T_1}{\pi M_1} = \frac{8 R T_2}{\pi M_2} \] The constant factor \(\frac{8R}{\pi}\) cancels out: \[ \frac{T_1}{M_1} = \frac{T_2}{M_2} \] We need to find the ratio \(\frac{T_2}{T_1}\). Let's rearrange the equation: \[ \frac{T_2}{T_1} = \frac{M_2}{M_1} \] Now, we use the given relationship between the molecular weights, \(M_1 = \frac{1}{2} M_2\). This implies that \(M_2 = 2 M_1\). Substituting this into our ratio: \[ \frac{T_2}{T_1} = \frac{2 M_1}{M_1} = 2 \] Step 4: Final Answer:
The ratio \(\frac{T_2}{T_1}\) is 2.