Problem 28 - Chemistry and Chemical Reactivity

Question

As the gas trapped in a cylinder with a movable piston cools, \(1.34 \mathrm{kJ}\) of work is done on the gas by the surroundings. If the gas is at a constant pressure of \(1.33 \times 10^{5} \mathrm{Pa}\), what is the change of volume (in L) of the gas?

Step-by-Step Solution

Answer

The change in volume is approximately \(10.08\, \text{L}\).

Step 1: Understand the given values

We are given the work done on the gas, which is \(1.34\, \text{kJ}\), and the pressure of the gas, which is \(1.33 \times 10^{5}\, \text{Pa}\). We need to determine the change in volume of the gas.

Step 2: Convert work to the standard unit

The work is given in kilojoules, so we convert it to joules because pressure is given in pascals (\(1\, \text{kJ} = 1000\, \text{J}\)). Thus, the work done on the gas is \(1.34 \times 1000 = 1340\, \text{J}\).

Step 3: Use the work formula

The work done on or by a gas at constant pressure is given by the formula: \(W = P \Delta V\), where \(W\) is the work, \(P\) is the pressure, and \(\Delta V\) is the change in volume. Since work is done on the gas, the formula becomes \(1340 = 1.33 \times 10^{5} \times \Delta V\).

Step 4: Solve for the change in volume \(\Delta V\)

Rearrange the formula to solve for \(\Delta V\): \[\Delta V = \frac{1340}{1.33 \times 10^{5}}\]

Step 5: Perform the calculation

Calculate the change in volume using the rearranged formula: \[\Delta V = \frac{1340}{133000} \approx 0.01007519\, \text{m}^3\]

Step 6: Convert the volume change to liters

Since \(1\, \text{m}^3 = 1000\, \text{L}\), convert the change in volume from cubic meters to liters: \(0.01007519\, \text{m}^3 \times 1000 = 10.07519\, \text{L}\).

Key Concepts

Work done by/on gas

In thermodynamics, understanding the concept of work done by or on a gas is crucial. The work done in this context refers to the energy transferred either from the surroundings to the gas or from the gas to the surroundings. In the exercise provided, work was done on the gas by the surroundings, meaning energy was transferred from the environment into the gas. This typically results in a compression of the gas, where its volume decreases under the influence of an external force.

The formula used to calculate work done in the case of a gas under constant pressure is:

• \( W = P \Delta V \)

Where:

• \( W \) is the work done,
• \( P \) is the pressure,
• \( \Delta V \) is the change in volume of the gas.

In our scenario, since the work done is 1.34 kJ (converted to 1340 J), and the pressure is constant, we can use this relationship to find the change in volume of the gas.

Constant pressure

When dealing with gases, the term "constant pressure" means that as the gas undergoes changes, its pressure remains the same. This simplification makes it easier to calculate other properties, such as work and change in volume. In practical scenarios, like the one in the exercise, the constant pressure assumption helps in direct application of the formula for work done by or on the gas.

Typically, when a gas is in a container with a movable piston, maintaining a constant pressure is possible as the piston adjusts its position. Here, the external force adjusts accordingly to ensure that even as the gas cools and its volume changes, the pressure does not vary. Understanding the condition of constant pressure is essential for accurately predicting the behavior of the gas during compression or expansion.

Constant pressure calculations often involve direct measurements of pressure, such as the 1.33 x 10⁵ Pa in our problem, which are crucial in determining the resulting changes as the system evolves.

Volume change in gases

The change in volume of a gas is a key factor when studying gas behaviors under different conditions. As temperature, pressure, or external forces change, the volume of a gas will adjust accordingly. In the provided exercise, the calculation focuses on finding out how the volume changes as work is done on the gas at a constant pressure.

By applying the equation \( W = P \Delta V \), we isolate \( \Delta V \) to find the change in volume, which is derived as follows:

• \( \Delta V = \frac{W}{P} \)

For our exercise, after substituting in the known values (after work conversion to Joules), we calculated the change in volume initially in cubic meters (0.01007519 m³) and then converted it to liters (10.07519 L).

This conversion is crucial because in practical applications, working with volumes in liters is often more intuitive and common.

Unit conversion in physics

Unit conversion is a fundamental part of physics that ensures measurements are accurate and comparable. In the given exercise, several unit conversions were necessary to solve the problem effectively. First, converting kilojoules to joules allowed us to match the units of work with those of pressure, ensuring consistent calculations.

Here's how the conversions are handled:

• Work: Convert from kJ to J: \( 1.34 \text{kJ} = 1.34 \times 1000 \text{J} = 1340 \text{J} \)
• Volume: Convert from m³ to L: \( 0.01007519 \text{m}^3 \times 1000 = 10.07519 \text{L} \)

These conversions rely on the basic unit relationships:

• 1 kJ = 1000 J
• 1 m³ = 1000 L

Understanding and executing these conversions are important skills in physics, ensuring that the results are presented in a meaningful and standardized way, which allows for easy interpretation and comparison.