Question

Consider two independent identical bulbs (A and B), each containing a gas. Bulb A has 2.00 moles of \(\mathrm{CH}_{4}\) and bulb \(\mathrm{B}\) has 2.00 moles of \(\mathrm{SO}_{2}\). These bulbs have a valve that can open into a long tube that has no gas (a vacuum). The tubes for each bulb are identical in length. (a) If both valves to bulbs \(\mathrm{A}\) and \(\mathrm{B}\) are opened simultaneously, which gas will reach the end of the tube first? (b) If both gases are to reach the end of the tube simultaneously, how would you alter the contents of each bulb? (You may not alter the bulbs or the length of the tube.)

Short answer

Answer: The gas CH4 reaches the end of the tube first. To make both gases reach the end simultaneously, we need to increase the temperature of Bulb B (SO2) without changing the temperature of Bulb A (CH4), following the equation: \(T_{B} = T_{A} \cdot \frac{M(\mathrm{SO}_{2})}{M(\mathrm{CH}_{4})}\).

Step-by-step solution

Recall the formula for root-mean-square speed

The root-mean-square (rms) speed of a gas is given by: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) where \(R\) is the ideal gas constant, \(T\) is the absolute temperature of the gas, and \(M\) is the molar mass of the gas.

Identify the molar mass of each gas

Given that Bulb A contains CH4 and Bulb B contains SO2, we first find their molar mass: M(CH4) = 12.01 g/mol (C) + 4 x 1.01 g/mol (H) = 16.05 g/mol M(SO2) = 32.07 g/mol (S) + 2 x 16.00 g/mol (O) = 64.07 g/mol

Analyze which gas will travel faster based on molar mass

Since both gases are released at the same temperature, the one with the lower molar mass travels faster. Because CH4 has a lower molar mass than SO2, it will have a higher root-mean-square speed and will travel faster through the tube. (a) Therefore, the gas that will reach the end of the tube first is CH4 from Bulb A.

Alter the contents of the bulbs to make the gases reach the end simultaneously

To make both gases reach the end of the tube simultaneously, we need to equate their root-mean-square speeds: \(v_{rms,\mathrm{CH}_4} = v_{rms,\mathrm{SO}_2}\) By substituting the formula for \(v_{rms}\), we get: \(\sqrt{\frac{3R(T_{A})}{M(\mathrm{CH}_{4})}} = \sqrt{\frac{3R(T_{B})}{M(\mathrm{SO}_{2})}}\) Since both bulbs have the same length of the tube and the ideal gas constant has the same value, we can simplify the equation as follows: \(\sqrt{\frac{T_{A}}{M(\mathrm{CH}_{4})}} = \sqrt{\frac{T_{B}}{M(\mathrm{SO}_{2})}}\) Now we can solve for \(T_{B}\), the temperature of Bulb B that ensures both gases reach the tube end simultaneously: \(T_{B} = T_{A} \cdot \frac{M(\mathrm{SO}_{2})}{M(\mathrm{CH}_{4})}\) (b) To make both gases reach the end of the tube simultaneously, we need to increase the temperature of Bulb B (SO2) without changing the temperature of Bulb A (CH4). The new temperature of Bulb B should satisfy the equation above.