Problem 109 - Chemistry The Central Science In Si Units

Question

An ideal gas at a pressure of \(152 \mathrm{kPa}\) is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of \(0.800 \mathrm{~L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is \(92.66 \mathrm{kPa}\), what is the volume of the bulb that was originally filled with gas?

Step-by-Step Solution

Answer

The volume of the originally filled bulb is approximately \(1.249\, \mathrm{L}\).

Step 1: Understand the Initial and Final States

In this problem, the gas initially at a pressure of \(152 \mathrm{kPa}\) is contained in a bulb of unknown volume \(V_1\). When it expands into a previously empty bulb of known volume \(V_2 = 0.800\, \mathrm{L}\), the final pressure becomes \(92.66 \mathrm{kPa}\). The temperature remains constant throughout the process.

Step 2: Use Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional: \( P_1 \cdot V_1 = P_2 \cdot (V_1 + V_2) \). This is because the initial product \(P_1V_1\) must equal the final product of pressure and volume \(P_2(V_1 + V_2)\) when temperature is constant.

Step 3: Substitute Known Values

Substitute the known pressures and volume into the Boyle's Law equation: \(152 \cdot V_1 = 92.66 \cdot (V_1 + 0.800)\). This substitution uses \(P_1 = 152\, \mathrm{kPa}\), \(P_2 = 92.66\, \mathrm{kPa}\), and \(V_2 = 0.800\, \mathrm{L}\).

Step 4: Solve for Unknown Volume \(V_1\)

Rearrange the equation from step 3 to solve for \(V_1\):\[152V_1 = 92.66V_1 + 92.66 \times 0.800\]First, distribute the \(92.66\) into the bracket:\[152V_1 = 92.66V_1 + 74.128\]Rearrange to solve for \(V_1\):\[59.34V_1 = 74.128\]Thus, \(V_1 = \frac{74.128}{59.34} = 1.249\, \mathrm{L}\).

Key Concepts

Ideal Gas

An ideal gas is a simplified model that helps us understand the behavior of gases under various conditions. It assumes that the gas particles are point particles with no volume and no intermolecular forces acting between them. This model is very useful for studying gas laws because it allows us to make predictions about gas behavior in different scenarios. In the context of this exercise, recognizing the gas as an ideal gas means we can apply simple relations like Boyle's Law without worrying about these additional complexities.

To consider gases as ideal, they should be at relatively high temperatures and low pressures, where the actual gas molecules are far enough apart that their size and interaction forces become negligible. While real gases deviate from this ideal behavior at very high pressures and low temperatures, many gases under normal conditions behave almost like ideal gases, making this concept extremely useful in simplifying calculations without introducing significant error.

Constant Temperature

The condition of constant temperature is one of the cornerstones of Boyle's Law, which is instrumental in solving this exercise. When we say the temperature is held constant, we refer to an isothermal process. In such processes, although pressure and volume can change, the temperature doesn't. This condition means any change in the volume and pressure of the gas doesn't result in a change in temperature, hence energy input or removal to maintain temperature isn't accounted for.

In our given problem, when the gas expands into a new bulb, the temperature remains constant, which allows us to directly apply Boyle's Law: \[ P_1 \times V_1 = P_2 \times V_2 \] Thus, knowing the temperature does not change assures us that it doesn't affect our equations and calculations for volumes and pressures in this scenario. Isothermal conditions are quite practical for theoretical studies since other factors influencing gas behavior can be ignored.

Volume Calculation

Volume calculation in an ideal gas scenario under constant temperature hinges on Boyle's Law: the inverse relationship between pressure and volume at a steady temperature. Our objective here is to find the unknown initial volume, denoted as \( V_1 \), of the bulb containing the gas.

Given the problem data, we know the initial pressure and final pressure as well as the volume the gas expands into:- P_1 = 152 \, \mathrm{kPa} - P_2 = 92.66 \, \mathrm{kPa}- V_2 = 0.800 \, \mathrm{L} Substituting these values into Boyle's Law gives us the equation:\[ 152V_1 = 92.66(V_1 + 0.800) \]By solving this equation, you can find:\[ 59.34V_1 = 74.128 \] Finally, the calculations yield the initial volume as:\[ V_1 = \frac{74.128}{59.34} \approx 1.249 \, \mathrm{L} \]This calculation reveals the original volume of the bulb before the gas expanded, illustrating the power of Boyle's Law under ideal conditions and constant temperature.