Problem 109 - Chemistry: The Central Science in SI Units, Global Edition

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 initial volume of the bulb filled with gas is approximately 0.583 L.

Step 1: Write down the initial and final states of pressure and volume

Before the stopcock is opened, the pressure is given as 152 kPa and the volume is unknown. We will denote the volume of the initial gas-filled bulb as V1. After the stopcock is opened, the final pressure is 92.66 kPa, and the total volume is the sum of the volume of the initial bulb (V1) and the evacuated bulb (0.800 L). We can denote the final volume as V2. Initial state: Pressure, P1 = 152 kPa, Volume, V1 (unknown) Final state: Pressure, P2 = 92.66 kPa, Volume, V2 = V1 + 0.800 L

Step 2: Use the ideal gas law to create an equation

Since the temperature is constant, we can write the ideal gas law for both the initial and final states, noting that the amount of gas, n, is also constant: Initial state: \(P_1V_1 = nRT\) Final state: \(P_2V_2 = nRT\)

Step 3: Solve for V1 using the conservation of volume principle

Divide the equation of the final state by the equation of the initial state: \(\frac{P_2V_2}{P_1V_1} = \frac{nRT}{nRT}\) The right side of the equation simplifies to 1, since n, R, and T are all constants: \(\frac{P_2V_2}{P_1V_1} = 1\) Now we can plug in the given values for P2, P1, and V2: \(\frac{92.66 (V_1 + 0.800)}{152 V_1} = 1\)

Step 4: Solve for V1

We can now solve for V1 by multiplying both sides by \(152V_1\): \(92.66 (V_1 + 0.800) = 152 V_1\) Divide both sides by 92.66: \(V_1 + 0.800 = \frac{152}{92.66} V_1\) Solve for V1: \(V_1(1 - \frac{152}{92.66}) = 0.800\) \(V_1 = \frac{0.800}{1 - \frac{152}{92.66}}\) Calculating the expression, we get: \(V_1 \approx 0.583 L\) So the initial volume of the bulb filled with gas is approximately 0.583 L.

Key Concepts

Pressure and Volume Relationship

The relationship between pressure and volume is one of the fundamental aspects of gas behavior described by the Ideal Gas Law. When analyzing gases, this law comes into play quite frequently. To understand the relationship, it's important to consider Boyle's Law, a specific case of the Ideal Gas Law.
The pressure of a gas is inversely proportional to its volume when temperature remains constant. This means:\[ P_1V_1 = P_2V_2 \]where:

• \( P_1 \) is the initial pressure,
• \( V_1 \) is the initial volume,
• \( P_2 \) is the final pressure, and
• \( V_2 \) is the final volume.

This inverse relationship implies that if the pressure of a gas increases, the volume must decrease, provided that the temperature is stable and vice versa. This is why, in the given exercise, as the gas expands into the previously empty bulb, a decrease in pressure is observed.

Gas Expansion

Gas expansion refers to how a gas can fill a larger volume when given the chance. In the context of the Ideal Gas Law, it is controlled by changes in pressure and volume under constant temperature scenarios. Imagine opening a door between two rooms - this is similar to opening a stopcock between two bulbs.
When the stopcock is opened in our exercise, the gas expands from its original bulb into the previously evacuated bulb. Since the temperature is kept constant, this becomes a perfect scenario for applying Boyle's Law. The number of gas molecules remains the same, but they now occupy a larger volume.
The result? Lower pressure, which is the outcome we aim for based on the initial and final conditions in the problem. Therefore, understanding gas expansion helps in predicting how pressure changes due to volume alterations as the gas fills the new space.

Combined Volume Calculation

When dealing with scenarios like the one in our exercise, it's essential to understand how to compute the combined volume after a gas expansion occurs. In this specific case, the total volume after the stopcock opens becomes crucial for calculations.
Here's how we approach it:

• Initially, you have an unknown volume \( V_1 \) connected to a bulb with known volume \( 0.800 L \).
• Upon opening the stopcock, the gas fills both bulbs, resulting in a combined volume \( V_2 \). Hence, \( V_2 = V_1 + 0.800 \).
• The final conditions help solve for the original unknown volume. Given that we know the initial and final pressures, and the combined volume from the expansion, using the formula \( \frac{P_2V_2}{P_1V_1} = 1 \) becomes instrumental.

From here, you solve for the unknown \( V_1 \), ensuring you can calculate how much volume the gas occupied originally, using present data. Understanding combined volume calculation is vital because it provides insights into how different volumes interact when a system undergoes expansion and pressure changes.