Problem 26 - Chemistry The Central Science In Si Units

Question

A fixed quantity of gas at \(25^{\circ} \mathrm{C}\) exhibits a pressure of \(99 \mathrm{kPa}\) and occupies a volume of \(4.00 \mathrm{~L}\). (a) Calculate the volume the gas will occupy if the pressure is increased to \(202.6 \mathrm{kPa}\) whilethe temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(100^{\circ} \mathrm{C}\) while the pressure is held constant.

Step-by-Step Solution

Answer

(a) The volume is 1.95 L; (b) The volume is 5.01 L.

Step 1: Understand the Given Situation

We are dealing with an ideal gas problem. The gas follows the ideal gas law, where the relationship between pressure, volume, and temperature is important. In this exercise, two scenarios are considered: (a) changing pressure while keeping temperature constant, and (b) changing temperature while keeping pressure constant.

Step 2: Identify Constants and Variables for (a)

For part (a), temperature is constant, and we need to find the volume when pressure changes. We use Boyle's Law which states: \( P_1V_1 = P_2V_2 \). Here, \(P_1 = 99\, \text{kPa}\), \(V_1 = 4.00\, \text{L}\), \(P_2 = 202.6\, \text{kPa}\), and \(V_2\) is unknown.

Step 3: Solve for Volume in (a)

Rearrange Boyle's Law to solve for \(V_2\): \( V_2 = \frac{P_1V_1}{P_2} \). Substitute the values to get: \( V_2 = \frac{99\, \text{kPa} \times 4.00\, \text{L}}{202.6\, \text{kPa}} \). Calculate \( V_2 \).

Step 4: Calculate V2 for (a)

Perform the calculation: \( V_2 = \frac{396}{202.6} \approx 1.95\, \text{L} \). This is the new volume when the pressure is increased to 202.6 kPa.

Step 5: Identify Constants and Variables for (b)

For part (b), pressure is constant and we need to find the volume when temperature changes. We use Charles's Law which states: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). Here, \(V_1 = 4.00\, \text{L}\), \(T_1 = 25^{\circ} \mathrm{C} = 298\, \text{K}\), \(T_2 = 100^{\circ} \mathrm{C} = 373\, \text{K}\), and \(V_2\) is unknown.

Step 6: Solve for Volume in (b)

Rearrange Charles's Law to solve for \(V_2\): \( V_2 = V_1 \times \frac{T_2}{T_1} \). Substitute the values to get: \( V_2 = 4.00 \times \frac{373}{298} \). Calculate \( V_2 \).

Step 7: Calculate V2 for (b)

Perform the calculation: \( V_2 = 4.00 \times 1.2517 \approx 5.01\, \text{L} \). This is the new volume when the temperature is increased to 100°C while keeping pressure constant.

Key Concepts

Boyle's Law

Boyle's Law provides a fascinating insight into how pressure and volume are intertwined for a fixed amount of gas at a constant temperature. Imagine compressing a balloon. As you squeeze it, the volume of gas inside decreases, while the pressure increases. This is exactly what Boyle's Law describes: when temperature is constant, the pressure of a gas varies inversely with its volume. Simply put, when one goes up, the other must come down. The law is mathematically expressed as:\[ P_1V_1 = P_2V_2 \]where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume respectively. This relationship is extremely helpful in predicting the behavior of gases when pressure changes are involved. By rearranging the equation, you can solve for a missing variable, such as when calculating the new volume a gas would occupy under a different pressure.

Charles's Law

Charles's Law reveals the direct relationship between temperature and volume for a gas at constant pressure. Picture a hot air balloon – as the air inside heats up, it expands, causing the balloon to rise. This principle is encapsulated in Charles's Law, which tells us that the volume of a gas increases with an increase in temperature, as long as the pressure is unchanged. The equation is given by:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]where \( V_1 \) and \( T_1 \) are the initial volume and temperature, and \( V_2 \) and \( T_2 \) are the final volume and temperature, with temperature measured in Kelvin. This formula allows us to solve for unknown variables and make predictions about how a gas's volume will adjust as its temperature changes.

Pressure and Volume Relationship

The relationship between pressure and volume is a fundamental concept in understanding gas behavior. It highlights how gas molecules behave based on changes in the space they occupy. When you increase the pressure on a gas by compressing it, its volume tends to decrease. However, if you release the pressure, the gas expands. This is an inverse relationship: as one variable increases, the other decreases. The explanation lies in the kinetic molecular theory, which posits that gas molecules move rapidly and are spread apart, with pressure resulting from their collisions with container walls. This relationship is crucial for various real-world applications, such as predicting how balloons will behave at different altitudes or designing systems that rely on gas compression, such as pumps and engines.

Temperature and Volume Relationship

The temperature and volume relationship is pivotal for understanding gas expansion and contraction in response to thermal energy changes. When you heat a gas, its molecules gain energy and move faster, thereby occupying more space. This explains why hot air balloons rise – the increased temperature expands the air inside, lowering the density of the balloon compared to the outside air. Conversely, cooling a gas decreases the speed of its molecules and causes the gas to shrink in volume. This relationship is direct: as one variable increases, so does the other. This concept is mirrored in real-world phenomena, from weather balloon behavior to the operation of internal combustion engines. It underlines the vital connection between a gas's thermal properties and its physical space requirements, providing insight into various practical applications and scientific explorations.