Question

A vessel has two equal compartments \(\mathrm{A}\) and \(\mathrm{B}\) containing \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\), respectively, each at 1 atm pressure. If the wall separating the compartment is removed, the pressure (1) Will remain unchanged in \(\mathrm{A}\) and \(\mathrm{B}\) (2) Will increase in \(\mathrm{A}\) and decrease in \(\mathrm{B}\) (3) Will decrease in \(\mathrm{A}\) and increase in \(\mathrm{B}\) (4) Will increase in both \(\mathrm{A}\) and \(\mathrm{B}\)

Short answer

The pressure in both compartments will remain unchanged at 1 atm after removing the wall.

Step-by-step solution

Understand the system

The vessel has two equal compartments, A containing \(\mathrm{H_{2}}\) and B containing \(\mathrm{O_{2}}\), both at 1 atm pressure. Initially, there is a wall separating the two compartments.

Apply Dalton's Law of Partial Pressures

When the separating wall is removed, the gases \(\mathrm{H_{2}}\) and \(\mathrm{O_{2}}\) will mix. According to Dalton's Law, the total pressure of the gas mixture is the sum of the partial pressures of each gas. Since each gas initially exerts a pressure of 1 atm in its compartment, after mixing, the combined pressure must be considered.

Determine the pressure changes

Both gases will now occupy the entire volume of the vessel. Each gas initially occupied half the volume of the vessel at 1 atm. When they mix and occupy the full volume, the partial pressure of each gas will be halved because pressure is inversely proportional to volume for a given number of moles and temperature (Boyle's Law). Thus, the partial pressure of \(\mathrm{H_{2}}\) will become 0.5 atm and the partial pressure of \(\mathrm{O_{2}}\) will also become 0.5 atm.

Calculate the total pressure

The total pressure in the vessel after removing the wall will be the sum of the partial pressures of \(\mathrm{H_{2}}\) and \(\mathrm{O_{2}}\). Therefore, the total pressure is \(0.5 + 0.5 = 1 \) atm.

Conclusion

Since the total pressure in the entire vessel remains the same (1 atm) and this pressure is equally distributed throughout the vessel, the pressure in both compartments A and B will remain unchanged at 1 atm when the wall is removed.

Key concepts

Gas Laws

Gas laws describe the behavior of gases and how they interact with temperature, volume, and pressure. There are several fundamental gas laws including Boyle's Law, Charles's Law, and Avogadro's Law. Understanding these laws helps us predict how gases will behave under different conditions.

According to the ideal gas law, the state of an ideal gas is defined by the equation: \[ PV = nRT \] where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.

In the context of the exercise, we are particularly looking at Dalton's Law of Partial Pressures which states that the total pressure exerted by a gas mixture is the sum of the partial pressures of each individual gas.

Partial Pressure

Partial pressure is a way to describe the contribution of each gas in a mixture to the total pressure. According to Dalton's Law of Partial Pressures, the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.

In the exercise, we have hydrogen (\(\text{H}_{2}\)) and oxygen (\(\text{O}_{2}\)) each initially at 1 atm in separate compartments. When the barrier is removed, these gases will mix. The partial pressure of each gas will adjust because they will now occupy a larger volume (the full volume of the vessel). Each gas will have its partial pressure halved to 0.5 atm, because pressure is inversely proportional to volume if temperature and mole count are constant.

Thus, the total pressure after mixing is the sum of the partial pressures: \[ P_{\text{total}} = P_{\text{H}_{2}} + P_{\text{O}_{2}} = 0.5 \text{ atm} + 0.5 \text{ atm} = 1 \text{ atm} \]

Boyle's Law

Boyle's Law describes how the pressure of a gas tends to decrease as the volume of the gas increases, assuming a constant temperature and a fixed amount of gas. The law can be expressed as: \[ P_{1}V_{1} = P_{2}V_{2} \] where \(P_{1}\) and \(V_{1}\) are the initial pressure and volume, and \(P_{2}\) and \(V_{2}\) are the final pressure and volume.

In the exercise, each gas starts with an initial state of 1 atm pressure in half of the volume of the vessel. When the barrier is removed, each gas expands to fill the entire volume. Boyle's Law helps us understand that as the volume doubles (since the gas moves into the other compartment), the pressure exerted by each gas will be halved. This is why the partial pressure of both \(\text{H}_{2}\) and \(\text{O}_{2}\) drops to 0.5 atm each.

Combining this with Dalton's Law helps us see that the total pressure remains unchanged, providing a clear and comprehensive understanding of the process described in the exercise.