Problem 35 - Chemistry An Atoms First Approach

Question

A sample of an ideal gas at \(15.0 \mathrm{atm}\) and \(10.0 \mathrm{L}\) is allowed to expand against a constant external pressure of \(2.00 \mathrm{atm}\) to a volume of \(75.0 \mathrm{L}\). Calculate the work in units of \(\mathrm{kJ}\) for the gas expansion.

Step-by-Step Solution

Answer

The work done by the gas expansion is approximately \(-13.17 \mathrm{kJ}\).

Step 1: Identify given values

We are given the following values: Initial pressure, P1 = 15.0 atm Initial volume, V1 = 10.0 L Final volume, V2 = 75.0 L Constant external pressure, P_ext = 2.00 atm

Step 2: Calculate the change in volume

To find the change in volume, subtract the initial volume from the final volume: ∆V = V2 - V1 ∆V = 75.0 L - 10.0 L ∆V = 65.0 L

Step 3: Convert pressure and volume to SI units

To calculate the work in joules, we need to convert the external pressure and change in volume to SI units. 1 atm = 101325 Pa 1 L = 0.001 m³ P_ext = 2.00 atm × 101325 Pa/atm = 202650 Pa ∆V = 65.0 L × 0.001 m³/L = 0.065 m³

Step 4: Calculate the work done

Now, use the formula for work done against constant pressure: Work = -P_ext × ∆V Work = -202650 Pa × 0.065 m³ Work = -13172.25 J

Step 5: Convert the work to kilojoules

To convert the work to kilojoules, divide by 1000: Work = -13172.25 J ÷ 1000 Work = -13.17 kJ The work done by the gas expansion is approximately -13.17 kJ.

Key Concepts

Ideal Gas

The concept of an ideal gas is an important foundation in thermodynamics. An ideal gas is a theoretical gas composed of a set of randomly-moving, non-interacting point particles. This idealization helps simplify calculations and predictions of how gases behave under various conditions.
In reality, no gas perfectly follows the ideal gas law, but many approximate it closely at high temperatures and low pressures. The ideal gas law is expressed as:

• \( PV = nRT \)

Here, \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
This equation provides a clear relationship between the pressure, volume, and temperature of a gas. Understanding and using this concept allows you to predict how a gas will behave when conditions change. In the context of the sample problem, treating the gas as ideal helps streamline the calculations.

Gas Expansion

Gas expansion occurs when a gas increases in volume due to changes in external conditions such as pressure or temperature. In this exercise, the gas expands from a volume of 10.0 liters to 75.0 liters.
The expansion of gas can be understood through various principles, such as Boyle's law, which states that pressure and volume have an inverse relationship when temperature is held constant.
This process can be isothermal, isobaric, adiabatic, or isochoric, depending on the constraints applied, such as keeping the temperature, pressure, heat, or volume constant respectively. In the given problem, the gas expansion is

• against a constant external pressure (2.00 atm).

This setup means the problem uses the principle of work done in an isobaric (constant pressure) process. As the volume increases, the gas does work on its surroundings, leading to a calculations of work done during the process.

Work Calculation

In thermodynamics, calculating work done by or on a system involves the energy required to change the volume of the gas. When gas expands against a constant external pressure, the work done can be calculated using the equation:

• \( ext{Work} = -P_{ ext{ext}} \times \Delta V \)

Where

• \( P_{ ext{ext}} \) is the constant external pressure, and
• \( \Delta V \) is the change in volume.

The negative sign indicates that the system (the gas) is doing work on the surroundings, meaning energy is leaving the system.
This is evident in the problem where the gas transition from 10.0 L to 75.0 L results in work done of \(-13.17\) kJ. This work is calculated after converting the values into consistent SI units.

SI Units Conversion

When working with physical quantities, consistent units are crucial to achieving correct results. In thermodynamics, quantities are often converted to SI units to streamline calculations. The problem at hand involves pressure and volume, which are initially given in atmospheres and liters respectively.To align with SI units, these are converted as follows:

• 1 atm is equivalent to 101325 pascals (Pa).
• 1 liter is equivalent to 0.001 cubic meters (m³).

For the given problem:

• External pressure is converted: 2.00 atm \( \times 101325 \text{ Pa/atm} = 202650 \text{ Pa} \).
• Change in volume is converted: 65.0 L \( \times 0.001 \text{ m}³/ ext{L} = 0.065 \text{ m}³ \).

Once converted, these values allow the calculation of work to adhere to SI standards, facilitating further analysis or comparisons.