Question

Two identical containers, one red and one yellow, are inflated with different gases at the same volume and pressure. Both containers have an identically sized hole that allows the gas to leak out. It takes four times as long for the yellow container to leak out compared to the red container. If the red container is twice as hot as the yellow container, what is the ratio of the molar masses of the gases (Myellow / Mred)

Answer 1

Answer:

8

Explanation:

Here we're dealing with the root mean square velocity of gases. We'll provide the formula in order to calculate the root mean square velocity of a gas:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

Here:

$R = 8.314 \frac{J}{K mol}$ is the ideal gas law constant;

$T$ is the absolute temperature in K;

$M$ is the molar mass of a compound in kg/mol.

We know that the gas from the red container is 4 times faster, as it takes 4 times as long for the yellow container to leak out, this means:

$\frac{v_{rms, red}}{v_{rms, yellow}} = 4$

We also know that the temperature of the red container is twice as large:

$\frac{T_{red}}{T_{yellow}} = 2$

Write the ratio of the velocities and substitute the variables:

$\frac{v_{rms, red}}{v_{rms, yellow}}=\frac{\sqrt{\frac{3RT_{red}}{M_{red}}}}{\sqrt{\frac{3RT_{yellow}}{M_{yellow}}}}=4$

Then:

$\frac{\sqrt{\frac{3RT_{red}}{M_{red}}}}{\sqrt{\frac{3RT_{yellow}}{M_{yellow}}}}=\sqrt{\frac{3RT_{red}}{M_{red}}\cdot \frac{M_{yellow}}{3RT_{yellow}}}=\sqrt{\frac{T_{red}}{T_{yellow}}\cdot \frac{M_{yellow}}{M_{red}}}=4$

From here:

$16 = \frac{T_{red}}{T_{yellow}}\cdot \frac{M_{yellow}}{M_{red}}$

Then:

$\frac{M_{yellow}}{M_{red}} = \frac{16}{\frac{T_{red}}{T_{yellow}}} = \frac{16}{2} = 8$