Problem 59 - Chemistry: An Atoms-Focused Approach

Question

Which of the following actions would produce the greater increase in the volume of a gas sample? a. Lowering the pressure from \(760 \mathrm{mm} \mathrm{Hg}\) to \(700 \mathrm{mmHg}\) at constant temperature b. Raising the temperature from \(10^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) at constant pressure

Step-by-Step Solution

Answer

The action that would produce a greater increase in the volume of the gas sample is raising the temperature from 10°C to 35°C at constant pressure. This is because the change in volume for this action (1.0884) is greater than the change in volume for decreasing the pressure from 760 mmHg to 700 mmHg at constant temperature (1.0857).

Step 1: Calculate initial volume of the gas sample

For this exercise, we can assume that we have a gas sample with fixed moles of gas, for comparison purposes. To calculate the initial volume before each action takes place, we can use the Ideal Gas Law formula: PV = nRT. Rearranging the formula, we get V = nRT/P For action a, the initial pressure is given as \(760\,\mathrm{mmHg}\). For action b, the initial temperature is given as \(10^{\circ}\mathrm{C}\). We must first convert this to Kelvin: $$T_{init} = 10^{\circ}\mathrm{C} + 273.15\, \mathrm{K} = 283.15\, \mathrm{K}$$

Step 2: Calculate final volume for each action

For action a, we are given a new pressure of \(700\,\mathrm{mmHg}\). Using the Ideal Gas Law formula V = nRT/P, and since the temperature is constant, we can calculate the ratio of the final volume to the initial volume: $$\frac{V_{f}}{V_{i}} = \frac{P_{i}}{P_{f}}$$ For action b, we are given a new temperature of \(35^{\circ}\mathrm{C}\). We must first convert this to Kelvin: $$T_{final} = 35^{\circ}\mathrm{C} + 273.15\, \mathrm{K} = 308.15\, \mathrm{K}$$ Using the Ideal Gas Law formula, and since the pressure is constant for action b, we can calculate the ratio of the final volume to the initial volume: $$\frac{V_{f}}{V_{i}} = \frac{T_{final}}{T_{init}}$$

Step 3: Compare the change in volume between both actions

Now we can calculate the change in volume for each action: For action a: $$\frac{V_{f}}{V_{i}} = \frac{760\,\mathrm{mmHg}}{700\,\mathrm{mmHg}} = 1.0857$$ For action b: $$\frac{V_{f}}{V_{i}} = \frac{308.15\,\mathrm{K}}{283.15\,\mathrm{K}} = 1.0884$$ From these calculations, we can see that the change in volume for action b (1.0884) is greater than the change in volume for action a (1.0857). Therefore, raising the temperature from \(10^{\circ}\mathrm{C}\) to \(35^{\circ}\mathrm{C}\) at constant pressure would produce a greater increase in the volume of the gas sample.

Key Concepts

Pressure-Volume Relationship

The pressure-volume relationship in gases is an intriguing concept described by Boyle's Law. This law tells us that the volume of a gas is inversely proportional to the pressure when the temperature remains constant. Imagine you have a balloon; as you squeeze it (increase the pressure), the volume decreases. Similarly, if you let go (decrease the pressure), the balloon expands, illustrating how gases behave.

Let's break down this idea further:

• Boyle's Law formula: \( P_1V_1 = P_2V_2 \)
• As pressure decreases, volume increases, provided temperature is constant.
• This is an inverse relationship because increasing one results in decreasing the other.

In our exercise, we observed such a change when the pressure dropped from 760 mmHg to 700 mmHg. The result? An increase in volume. The decrease in pressure allowed the gas particles more space to move, thus increasing the volume.

Temperature-Volume Relationship

The temperature-volume relationship is encapsulated by Charles's Law, stating that the volume of a gas is directly proportional to its temperature, as long as the pressure is kept unchanged. To visualize, think of a hot air balloon. As it gets heated, the air expands, causing the balloon to rise.

Consider these points to better grasp this relationship:

• Charles's Law formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
• If the temperature rises, the volume increases proportionally, assuming constant pressure.
• Always convert Celsius to Kelvin before performing calculations for accuracy.

The exercise demonstrates this with a temperature increase from 10°C to 35°C, converted to Kelvin. This significantly impacted the volume, as the gas expanded due to the increased kinetic energy of the particles.

Gas Laws Calculations

Understanding gas laws calculations involves appreciating how the Ideal Gas Law bridges all relationships in gases. The formula \( PV = nRT \) is instrumental in solving many gas behavior scenarios, linking pressure, volume, temperature, and moles (n).

This is useful when:

• Comparing initial and final states of gases in terms of pressure, volume, and temperature changes.
• Determining one variable when the others are known.
• In practice, can help verify results from Boyle's and Charles's Laws.

In our exercise, we precisely computed changes in volume due to alterations in pressure and temperature by manipulating the Ideal Gas Law for each action. These calculations gave us clear comparative insights into how different actions affected the volume, showcasing the comprehensive utility of the Ideal Gas Law in conceptual understanding and real-world applications.