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

STEPS:

1. Identify the Initial Conditions and Variables: The problem provides two separate gas samples, Helium ($He$) and Nitrogen ($N_2$), both at a temperature of $0^\circ\text{C}$ ($273\text{ K}$) and a pressure of $1.0\text{ atm}$. They are moved into a rigid $4.0\text{ L}$ container, and the temperature remains constant at $0^\circ\text{C}$.
2. Determine the Relevant Gas Law: Since temperature and the total number of moles of gas remain constant, Boyle's Law ($P_1V_1 = P_2V_2$) and Dalton’s Law of Partial Pressures ($P_{total} = P_A + P_B + \dots$) are the primary concepts tested.
3. Calculate the Partial Pressure of Helium: Treat the Helium independently to find its pressure in the new $4.0\text{ L}$ volume:
* $P_1 = 1.0\text{ atm}$, $V_1 = 6.0\text{ L}$, $V_2 = 4.0\text{ L}$
* $(1.0\text{ atm})(6.0\text{ L}) = (P_{He})(4.0\text{ L})$
* $P_{He} = 1.5\text{ atm}$
4. Calculate the Partial Pressure of Nitrogen: Repeat the process for the Nitrogen:
* $P_1 = 1.0\text{ atm}$, $V_1 = 10.\text{ L}$, $V_2 = 4.0\text{ L}$
* $(1.0\text{ atm})(10.\text{ L}) = (P_{N2})(4.0\text{ L})$
* $P_{N2} = 2.5\text{ atm}$
5. Calculate the Final Total Pressure: According to Dalton's Law, the total pressure in the container is the sum of the partial pressures:
* $P_{total} = 1.5\text{ atm} + 2.5\text{ atm} = \mathbf{4.0\text{ atm}}$
* Alternative Method: One can also find the total "volume" at $1\text{ atm}$ first ($6.0\text{ L} + 10.\text{ L} = 16\text{ L}$) and then apply Boyle's Law to the mixture: $(1.0\text{ atm})(16\text{ L}) = (P_{final})(4.0\text{ L})$, which also yields $4.0\text{ atm}$.

WHY_OTHERS_WRONG:

• A) 2.0 atm: This value does not result from a standard application of gas laws to the given volumes. A student might arrive at this through a major calculation error or by incorrectly averaging the individual partial pressures.
• C) 6.4 atm: This might result from an incorrect calculation, such as attempting to use the ratio of volumes in a way that squares the denominators or applying a proportionality constant incorrectly.
• D) 8.8 atm: There is no simple path to this number using the variables provided in the problem.
• E) 16 atm: This value represents the total volume in liters of the gases if they were kept at the initial pressure of $1.0\text{ atm}$. It is the $P \times V$ product ($16\text{ L} \cdot \text{atm}$), but it is not the pressure once that gas is compressed into a $4.0\text{ L}$ space.