Question

Consider three sealed tanks all at the same temperature, pressure, and volume. Tank A contains \(\mathrm{SO}_{2}\) gas. Tank B contains \(\mathrm{O}_{2}\) gas. Tank C contains \(\mathrm{CH}_{4}\) gas. Use LT (for "is less than"), GT (for "is greater than"), EQ (for "is equal to"), or MI (for " more information required") as answers to the blanks below. (a) The mass of \(\mathrm{SO}_{2}\) in tank \(\mathrm{A}\) the mass of \(\mathrm{O}_{2}\) in \(\operatorname{tank} \mathrm{B}\). (b) The average translational energy of \(\mathrm{CH}_{4}\) in \(\operatorname{tank} \mathrm{C}\) the average translational energy of \(\mathrm{SO}_{2}\) in \(\operatorname{tank} \mathrm{A}\) (c) It takes \(20 \mathrm{~s}\) for all of the \(\mathrm{O}_{2}\) gas in tank \(\mathrm{B}\) to effuse out of a pinhole in the tank. The time it takes for all of the \(\mathrm{SO}_{2}\) to effuse out of tank A from an identical pinhole \(40 \mathrm{~s}\) (d) The density of \(\mathrm{O}_{2}\) in tank \(\mathrm{B} \longrightarrow\) the density of \(\mathrm{CH}_{4}\) in tank C. (e) The temperature in tank \(\mathrm{A}\) is increased from \(150 \mathrm{~K}\) to \(300 \mathrm{~K}\). The temperature in tank \(\mathrm{B}\) is kept at \(150 \mathrm{~K}\). The pressure in tank \(\mathrm{A}\) is half the pressure in \(\operatorname{tank}\) B.

Short answer

Answer: MI (more information needed)

Step-by-step solution

Calculate the molar masses

To compare the mass, we need to calculate the molar mass of each of these gases. We know that the molar mass of S = 32 g/mol, O = 16 g/mol, and H = 1 g/mol. So, the molar mass of \(\mathrm{SO}_{2}\) is (32 + 2 * 16) g/mol, and the molar mass of \(\mathrm{O}_{2}\) is (2 * 16) g/mol.

Compare the molar masses of gases

With the molar masses calculated, we find that the molar mass of \(\mathrm{SO}_{2}\) is greater than the molar mass of \(\mathrm{O}_{2}\). Since all tanks have the same volume, temperature, and pressure, they contain the same number of moles, according to the ideal gas law (\(PV=nRT\)). Therefore, since the molar mass of \(\mathrm{SO}_{2}\) > molar mass of \(\mathrm{O}_{2}\), we can infer that the mass of \(\mathrm{SO}_{2}\) in tank A > mass of \(\mathrm{O}_{2}\) in tank B. The correct answer is GT. (b) The average translational energy of \(\mathrm{CH}_{4}\) in tank C the average translational energy of \(\mathrm{SO}_{2}\) in tank A

Assess the information about gases in tanks A and C

Since tanks A and C have the same temperature, pressure, and volume, their average translational energy will be the same as they follow the same Ideal Gas Law with the same conditions. The correct answer is EQ. (c) The time it takes for all of the \(\mathrm{SO}_{2}\) to effuse out of tank A from an identical pinhole \(40 \mathrm{~s}\)

Apply the Graham's law of effusion

Graham's law of effusion states: \(\frac{t_{1}}{t_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}\), where t1 and t2 are the time taken for two gases, and M1 and M2 are their molar masses. We have given the time \(\mathrm{O}_{2}\) takes to effuse, and we're asked to compare this with the time for \(\mathrm{SO}_{2}\) to effuse.

Compare the effusion times for both gases

To calculate the ratio, insert the molar masses \(\frac{t_{\mathrm{SO}_{2}}}{t_{\mathrm{O}_{2}}} = \sqrt{\frac{\text{Molecular weight of }\mathrm{O}_{2}}{\text{Molecular weight of }\mathrm{SO}_{2}}} = \sqrt{\frac{32}{64}} = \frac{1}{\sqrt{2}}\). Since we know that the time taken for \(\mathrm{O}_{2}\) to effuse is 20 seconds, we can find the time taken for \(\mathrm{SO}_{2}\) to effuse: \(t_{\mathrm{SO}_{2}}=20\sqrt{2}\). The answer is MI, since we cannot directly compare the given value of 40 seconds with the calculated value of \(20\sqrt{2}\) seconds. (d) The density of \(\mathrm{O}_{2}\) in tank B the density of \(\mathrm{CH}_{4}\) in tank C

Calculate the density of the gases

To determine the densities, we need to get the molar mass of \(\mathrm{CH}_{4}\) which is (12 +1(4)) g/mol. The density of a gas can be calculated as: Density = \(\frac{\text{mass}}{V} = \frac{\text{n} \times \text{ Molar mass}}{V} = \frac{\frac{P \times V}{R \times T}\times \text{ Molar mass}}{V} = \frac{P \times \text{ Molar mass}}{R \times T}\), where P represents pressure, V represents volume, R represents the gas constant, and T represents the temperature. Since A, B, and C have the same pressure, volume, and temperature, we can compare their densities by just comparing their molar masses.

Compare the densities>

Comparing the molar masses of the gases, we find that the molar mass of \(\mathrm{O}_{2} > \mathrm{CH}_{4}\). This means that the density of \(\mathrm{O}_{2}\) in tank B > the density of \(\mathrm{CH}_{4}\) in tank C. The answer is GT. (e) The pressure in tank A is half the pressure in tank B

Apply the ideal gas law for tank A

For tank A, the ideal gas law is: \(P_{A}V_{A}=n_{A}RT_{A}\). After the temperature is increased in tank A, the new ideal gas law will be \(P_{A'}V_{A'}=n_{A}R(T_{A}+150)\). The volume and the number of moles do not change.

Apply the ideal gas law for tank B

For tank B, the ideal gas law is \(P_{B}V_{B}=n_{B}RT_{B}\). In this case, the temperature was kept constant at 150K.

Compare the pressures

We have been given the relation \(P_A = \frac{1}{2} P_B\). If we compare the pressures of tanks A and B: \(\frac{P_A}{P_B} = \frac{n_{A}R(T_{A}+150)}{n_{B}RT_{B}} = \frac{1}{2}\). The temperature in B is kept at 150 K, and in A, it is increased from 150 K to 300 K. Thus, the ratio of temperatures would require more information to confirm if the pressure in tank A after the temperature increase is half the pressure in tank B as more information is needed about the initial pressures and the number of moles in the tanks. The answer is MI.