Problem 30 - Pearson Physics

Question

Calculate An ideal gas is compressed at a constant pressure of \(120 \mathrm{kPa}\) to one-half of its initial volume. The work done on the gas is \(790 \mathrm{~J}\). What was the initial volume of the gas?

Step-by-Step Solution

Answer

The initial volume of the gas was 13.167 liters.

Step 1: Understand the Problem

We have an ideal gas compressed at constant pressure, and the work done on the gas is given. We need to determine the initial volume of the gas.

Step 2: Identify the Given Values

The constant pressure is given as \( P = 120 \text{ kPa} \). The work done on the gas is \( W = 790 \text{ J} \). The final volume is half of the initial volume, so if the initial volume is \( V_i \), the final volume is \( V_f = \frac{V_i}{2} \).

Step 3: Recall the Work Formula for Constant Pressure

The formula for work done on an ideal gas at constant pressure is given by: \[ W = P (V_f - V_i) \] where \( V_f \) is the final volume and \( V_i \) is the initial volume.

Step 4: Substitute Known Values

Substitute the known values from Step 2 into the formula:\[ 790 = 120 (\frac{V_i}{2} - V_i) \] Remember to convert 120 kPa to Pa: \( 120 \text{ kPa} = 120,000 \text{ Pa} \).

Step 5: Simplify the Equation

Substitute the converted pressure into the equation:\[ 790 = 120,000 \left( \frac{V_i}{2} - V_i \right) \] Simplify the expression inside the parenthesis: \[ \frac{V_i}{2} - V_i = -\frac{V_i}{2} \] Thus, the equation becomes:\[ 790 = 120,000 \times (-\frac{V_i}{2}) \]

Step 6: Solve for Initial Volume \( V_i \)

Rearrange the equation to solve for \( V_i \):\[ -790 = -60,000 \times V_i \]Divide both sides by \(-60,000\): \[ V_i = \frac{790}{60,000} \] Calculate to find \( V_i \).

Step 7: Calculate the Initial Volume

Compute the division: \( V_i = \frac{790}{60,000} = 0.0131667 \) cubic meters or \( 13.167 \) liters (since 1 m³ = 1000 liters).

Key Concepts

Constant Pressure

When dealing with ideal gases, one common scenario is when the gas is compressed or expanded under constant pressure. This means that while the volume of the gas may change, the pressure remains the same throughout the process.
In such cases, it becomes easier to apply certain mathematical formulas, because pressure remains a constant known value.
This situation is commonly described using a horizontal line on a pressure-volume (PV) diagram. Keeping pressure constant can often simplify calculations and understanding how a gas behaves when its volume or temperature changes.

• This constant value is often specified in units like Pascals (Pa) or kilopascals (kPa) for convenience.
• In this exercise, the gas is compressed at a constant pressure of 120 kPa.

Understanding and handling constant pressure processes is crucial in thermodynamics and helps us solve problems like predicting volumes or calculating work done.

Initial Volume

The initial volume of a gas is a critical parameter when exploring gas laws. It represents the volume that a gas occupies before any changes in pressure, temperature, or volume occur.
In the context of the ideal gas equation, the initial volume is often what we aim to determine or what us provided as a starting point for calculations. A clear understanding of initial conditions, like the initial volume, is essential because it allows us to accurately predict how the gas will behave when external changes are applied.
In our exercise, the initial volume is unknown, but we know that the final volume is half of the initial volume. Once we understand the relationship between initial and final volumes, we can use various thermodynamic equations to describe the behavior of the gas.

• In this exercise, calculating the initial volume involves working backward using known values like pressure and the work done on the gas.

This reverse calculation helps us appreciate the dependency and influence of each variable in thermodynamic processes.

Work Done on Gas

The concept of work done on gas in thermodynamics is related to how energy is transferred either into or out of the system (in this case, the ideal gas).
The work done is a measure of energy that is required to change the volume of the gas under certain conditions, such as constant pressure.In general, the work done on or by a gas during a process is calculated using the equation:
\[ W = P (V_f - V_i) \]Here,

• \(W\) represents the work done (in joules),
• \(P\) is the pressure (held constant here),
• \(V_f\) and \(V_i\) are the final and initial volumes respectively.

In our exercise, we used this equation to find the initial volume given the work done (790 J) and the fact that the volume is halved at a constant pressure of 120 kPa.The work done on the gas reflects the energy transfer necessary for the volume change.
By understanding this relationship, we can solve and analyze real-world problems scientifically and practically.