Problem 3 - Chemistry: The Central Science

Question

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of 2\(?[\) Section 10.3\(]\) \begin{equation}\begin{array}{l}{\text { (a) It would contain the same number of molecules. }} \\ {\text { (b) It would contain half as many molecules. }} \\\ {\text { (c) It would contain twice as many molecules. }} \\ {\text { (d) There is insufficient data to say. }}\end{array}\end{equation}

Step-by-Step Solution

Answer

(b) It would contain half as many molecules.

Step 1: Apply Boyle's Law

As the volume V and the temperature T remain constant while the pressure P changes, we can apply Boyle's Law. Boyle's Law states that the product of the pressure and volume (P × V) remains constant for an ideal gas when held at a constant temperature. Mathematically, it can be represented as: \(P_1V_1 = P_2V_2\) where \(P_1\) and \(P_2\) are the initial and final pressures, and \(V_1\) and \(V_2\) are the initial and final volumes.

Step 2: Use the given condition to find the relationship between the initial and final pressures

Given that the pressure decreases by a factor of 2, we can relate the initial and final pressure as: \(P_2 = \dfrac{P_1}{2}\) Since the volume remains constant, we can say that \(V_1 = V_2\).

Step 3: Determine the relationship between the initial and final number of molecules

Using the Ideal Gas Law, we can find the relationship between the initial and final number of molecules of the gas. The Ideal Gas Law is expressed as: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. For the initial state: \(P_1V_1 = n_1RT_1\) For the final state: \(P_2V_2 = n_2RT_2\) As temperature and volume remain constant, we have: \(T_1 = T_2\) and \(V_1 = V_2\). Since \(P_1V_1 = P_2V_2\), we can write: \(n_1RT_1 = n_2RT_2\) Dividing both sides by RT: \(\dfrac{n_1T_1}{n_2T_2} = \dfrac{n_1}{n_2}\) As T is constant, we have: \(\dfrac{n_1}{n_2} = \dfrac{P_1V_1}{P_2V_2}\) Since the volume is constant, we can write the relation as: \(\dfrac{n_1}{n_2} = \dfrac{P_1}{P_2}\) Now substituting the value of \(P_2\) from step 2: \(\dfrac{n_1}{n_2} = \dfrac{P_1}{\frac{P_1}{2}}\) \(\dfrac{n_1}{n_2} = 2\) Here, \(n_2 = \dfrac{n_1}{2}\) which means there are now half as many molecules.

Step 4: Choose the correct answer based on the analysis

Based on the relationship between the initial and final number of molecules, the correct answer is: (b) It would contain half as many molecules.