When 4.0 L of He(g), 6.0 L of N2(g), and 10. L of Ar(g), all at 0°C and 1.0 atm, are pumped into an — States of Matter Chemistry Question

STEPS:

1. Identify the relevant physical laws and constants: Because the temperature of all gases remains constant at $0^\circ\text{C}$ throughout the entire process, we can analyze this system using two key gas principles:
* Boyle's Law ($P_1V_1 = P_2V_2$): Tells us that the pressure of a gas is inversely proportional to its volume when temperature and moles are constant.
* Dalton's Law of Partial Pressures ($P_{\text{total}} = P_1 + P_2 + P_3 + \dots$): States that the total pressure of a mixture of non-reacting gases is equal to the sum of the individual partial pressures of each gas in the container.

2. Calculate the partial pressure of Helium ($\text{He}$) in the new container:
Helium starts at a volume of $4.0\text{ L}$ and a pressure of $1.0\text{ atm}$. When pumped into the $8.0\text{ L}$ container:
$$P_{\text{initial}} V_{\text{initial}} = P_{\text{final}} V_{\text{final}}$$
$$(1.0\text{ atm})(4.0\text{ L}) = P_{\text{He}}(8.0\text{ L}) \implies P_{\text{He}} = \frac{4.0\text{ L} \cdot \text{atm}}{8.0\text{ L}} = \mathbf{0.50\text{ atm}}$$

3. Calculate the partial pressure of Nitrogen ($\text{N}_2$) in the new container:
Nitrogen starts at a volume of $6.0\text{ L}$ and a pressure of $1.0\text{ atm}$. When pumped into the $8.0\text{ L}$ container:
$$(1.0\text{ atm})(6.0\text{ L}) = P_{\text{N}_2}(8.0\text{ L}) \implies P_{\text{N}_2} = \frac{6.0\text{ L} \cdot \text{atm}}{8.0\text{ L}} = \mathbf{0.75\text{ atm}}$$

4. Calculate the partial pressure of Argon ($\text{Ar}$) in the new container:
Argon starts at a volume of $10.\text{ L}$ and a pressure of $1.0\text{ atm}$. When pumped into the $8.0\text{ L}$ container:
$$(1.0\text{ atm})(10.\text{ L}) = P_{\text{Ar}}(8.0\text{ L}) \implies P_{\text{Ar}} = \frac{10.\text{ L} \cdot \text{atm}}{8.0\text{ L}} = \mathbf{1.25\text{ atm}}$$

5. Sum the partial pressures to find the total final pressure:
Using Dalton's Law, add the newly calculated partial pressures of the three gases together:
$$P_{\text{total}} = P_{\text{He}} + P_{\text{N}_2} + P_{\text{Ar}}$$
$$P_{\text{total}} = 0.50\text{ atm} + 0.75\text{ atm} + 1.25\text{ atm} = \mathbf{2.5\text{ atm}}$$

*Alternative Mole^-Proportion Method (Mental Math/Short Cut):*
* According to Avogadro’s Law, under the same conditions of temperature ($0^\circ\text{C}$) and pressure ($1.0\text{ atm}$), volume is directly proportional to the number of moles of gas.
* The total equivalent volume of the separate gases combined is:
$$4.0\text{ L} + 6.0\text{ L} + 10.\text{ L} = \mathbf{20.\text{ L of gas at 1.0 atm}}$$
* Compressing this total quantity of gas from its equivalent original volume of $20.\text{ L}$ down to an $8.0\text{ L}$ rigid container allows us to apply Boyle's Law directly to the combined mixture:
$$P_1V_1 = P_2V_2 \implies (1.0\text{ atm})(20.\text{ L}) = P_{\text{final}}(8.0\text{ L}) \implies P_{\text{final}} = \frac{20.}{8.0} = \mathbf{2.5\text{ atm}}$$
* This confirms Option C is the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (0.5 atm): This value is only the partial pressure of Helium in the final container. A student would make this error if they performed the calculations for the first gas but forgot to account for the contributions of the nitrogen and argon gases.
• Option B is incorrect (1.0 atm): This is the original starting pressure of each gas. This would only be the final pressure if the total volume of the combined gases ($20.\text{ L}$) remained unchanged. Because the gases were compressed into a much smaller space ($8.0\text{ L}$), the pressure must increase.
• Option D is incorrect (4.0 atm): This option could result from a student attempting to divide the volume of Helium ($4.0\text{ L}$) by its pressure, or from incorrect arithmetic where they assume the pressures of the three systems are simply additive ($1.0 + 1.0 + 1.0 = 3.0\text{ atm}$) and then attempt to add a volume factor.