Questions 14-17 refer to the following information. Substance | Lewis Diagram | Boiling Point CH3OH — States of Matter Chemistry Question

STEPS:

1. Understand the state of the substance at 400 K:
The problem states that when the temperature of $\text{CH}_3\text{OH}$ is increased to $400\text{ K}$, all of the liquid vaporizes. This is a crucial observation: because there is no liquid remaining, the system is no longer in a liquid-vapor phase equilibrium. Instead, it is a single-phase gas system governed solely by the gas laws.

2. Relate the mass of the sample to the number of moles of gas:
The number of moles of a substance ($n$) is calculated by dividing its mass ($m$) by its molar mass ($M$):
$$n = \frac{m}{M}$$
Because the identity of the chemical ($\text{CH}_3\text{OH}$) does not change, its molar mass is constant. Therefore, repeating the experiment with half the mass means that the container now contains exactly half the number of moles of gaseous methanol ($n_{\text{new}} = 0.5 \times n_{\text{original}}$).

3. Apply the Ideal Gas Law to the constant-volume system:
The relationship between pressure, volume, temperature, and moles of a gas is given by the ideal gas equation:
$$PV = nRT \Rightarrow P = n \times \left(\frac{RT}{V}\right)$$
* The volume of the rigid vessel is constant ($V = 2.0\text{ L}$).
* The final temperature is constant ($T = 400\text{ K}$).
* The gas constant ($R$) is constant.
* Since the term $\frac{RT}{V}$ is entirely constant, the pressure of the gas is directly proportional to the number of moles of gas present ($P \propto n$).

4. Calculate the new pressure:
Because all of the methanol will also completely vaporize under these conditions (with even less mass, there is no liquid remaining to limit the pressure), we can directly scale the pressure based on the proportional decrease in moles:
$$P_{\text{new}} = 0.5 \times P_{\text{original}}$$
$$P_{\text{new}} = 0.5 \times 0.30\text{ atm} = \mathbf{0.15\text{ atm}}$$
This matches Option A.

*

WHY_OTHERS_WRONG:

• Option B is incorrect (0.30 atm): A student might choose this if they mistakenly assume that the pressure at $400\text{ K}$ represents an equilibrium vapor pressure. If the system were at liquid-vapor equilibrium (where liquid is still present on the bottom), the vapor pressure would indeed remain constant regardless of the amount of liquid. However, because *all* the liquid vaporized in the original experiment, the pressure is determined by gas stoichiometry, which drops when the quantity of matter is halved.
• Option C is incorrect (0.40 atm): This value is larger than the original pressure. Halving the amount of gas in a fixed volume at a constant temperature cannot result in an increased pressure.
• Option D is incorrect (0.60 atm): This is double the original pressure. This would be the pressure if the mass of the methanol had been *doubled* instead of halved, representing a misinterpretation of the direct relationship between moles and pressure.