Problem 17 - Pearson Physics

Question

As a gas expands at constant pressure from a volume of \(0.74 \mathrm{~m}^{3}\) to a volume of \(2.3 \mathrm{~m}^{3}\), it does \(93 \mathrm{~J}\) of work. What is the pressure of the gas during this process?

Step-by-Step Solution

Answer

The pressure of the gas during the process is approximately \( 59.62 \mathrm{~Pa} \).

Step 1: Understanding Work Done by Gas

When a gas expands at a constant pressure, it can perform work. The work done, denoted as \( W \), is given by the formula: \[ W = P \Delta V \] where \( P \) is the pressure and \( \Delta V \) is the change in volume of the gas.

Step 2: Calculate Change in Volume

The change in volume, \( \Delta V \), can be calculated by subtracting the initial volume from the final volume: \[ \Delta V = V_f - V_i \] \( V_f = 2.3 \mathrm{~m}^3 \) and \( V_i = 0.74 \mathrm{~m}^3 \), therefore: \[ \Delta V = 2.3 - 0.74 = 1.56 \mathrm{~m}^3 \]

Step 3: Rearrange Formula to Solve for Pressure

Since we know the work done, \( W = 93 \mathrm{~J} \), and the change in volume, \( \Delta V = 1.56 \mathrm{~m}^3 \), we can rearrange the work formula to solve for pressure: \[ P = \frac{W}{\Delta V} \]

Step 4: Calculate the Pressure

Substituting the known values into the equation: \[ P = \frac{93 \mathrm{~J}}{1.56 \mathrm{~m}^3} = 59.615 \mathrm{~Pa} \] This is the pressure of the gas during the expansion process.

Key Concepts

Constant Pressure Process

In thermodynamics, a constant pressure process refers to a situation where the pressure remains unchanged while the volume of a gas changes. This condition simplifies calculations because the pressure is a known value. When dealing with processes like gas expansion, it's vital to understand the behavior under constant pressure conditions because it directly ties into how work is done on or by the gas.

During a constant pressure expansion, the work done is directly proportional to the change in volume. In more straightforward terms, as a gas expands at a constant pressure, it needs energy to push against external factors, often resulting in work done by the gas. This work can be calculated using the formula:

• Work Done (\(W\)) = Pressure (\(P\)) \(\times\) Change in Volume (\(\Delta V\))

This formula allows us to see the linear relationship between pressure and volume change. So, the larger the volume the gas expands to, the more work it performs, assuming the pressure remains constant.

Change in Volume

The change in volume of a gas during an expansion or compression is crucial in calculating various parameters, including the work done by the gas. It is represented by the symbol \(\Delta V\), which symbolizes the difference between the final and initial volumes. Essentially, it tells us how much the size of the gas container has changed.

To compute \(\Delta V\), you subtract the initial volume (\(V_i\)) from the final volume (\(V_f\)):

• \(\Delta V = V_f - V_i\)

In our example, with a gas expanding from \(0.74\, \mathrm{m}^3\) to \(2.3\, \mathrm{m}^3\), the change in volume is \(1.56\, \mathrm{m}^3\). This change is what determines the amount of space the gas now occupies compared to its original state. Understanding and measuring \(\Delta V\) accurately is pivotal, as it feeds directly into calculations for work and pressure changes.

Pressure Calculation

Pressure is a critical parameter in thermodynamics. When a gas expands at constant pressure and performs a known amount of work, it allows us to backtrack and determine the pressure exerted by the gas if not explicitly given.

In a constant pressure process, the formula for work, \(W = P \Delta V\), can be rearranged to solve for pressure (\(P\)):

• \(P = \frac{W}{\Delta V}\)

By plugging in the work done and the change in volume, you can easily calculate the pressure. For instance, if the work done is \(93\, \mathrm{J}\) and the change in volume is \(1.56\, \mathrm{m}^3\), the pressure of the gas would be:

• \(P = \frac{93 \mathrm{~J}}{1.56 \mathrm{~m}^3} = 59.615 \mathrm{~Pa}\)

This calculation is straightforward because the work-rounding relationship helps obtain the pressure with a simple division, providing a tangible sense of the pressure level during the gas's expansion.