Question

A gas expands in volume from 26.7 to \(89.3 \mathrm{~mL}\) at constant temperature. Calculate the work done (in joules) if the gas expands (a) against a vacuum, (b) against a constant pressure of \(1.5 \mathrm{~atm},\) and \((\mathrm{c})\) against a constant pressure of \(2.8 \mathrm{~atm} .(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J})\).

Short answer

(a) 0 J, (b) -9.51 J, (c) -17.76 J

Step-by-step solution

Identify the Known Variables

Before calculating the work done, identify the known variables in the problem:- Initial volume, \( V_i = 26.7 \) mL (or 0.0267 L after converting to liters)- Final volume, \( V_f = 89.3 \) mL (or 0.0893 L after converting to liters)- Conversion factor: \( 1 \text{ L} \cdot \text{ atm} = 101.3 \text{ J} \)

Calculate Work for Expansion Against a Vacuum

When a gas expands against a vacuum, there is no external pressure opposing the expansion, so the work done by the gas is zero:\[ W = 0 \text{ J} \]

Calculate Work Against a Constant Pressure of 1.5 atm

Using the formula for work done by gas expansion, \( W = -P \Delta V \), where \( \Delta V = V_f - V_i \), we have:\[ \Delta V = 0.0893 \text{ L} - 0.0267 \text{ L} = 0.0626 \text{ L} \]Substituting into the formula:\[ W = -1.5 \text{ atm} \times 0.0626 \text{ L} = -0.0939 \text{ L} \cdot \text{ atm} \]Converting to joules:\[ W = -0.0939 \text{ L} \cdot \text{ atm} \times 101.3 \text{ J/L} \cdot \text{ atm} = -9.51 \text{ J} \]

Calculate Work Against a Constant Pressure of 2.8 atm

Again using \( W = -P \Delta V \):\[ W = -2.8 \text{ atm} \times 0.0626 \text{ L} = -0.17528 \text{ L} \cdot \text{ atm} \]Converting to joules:\[ W = -0.17528 \text{ L} \cdot \text{ atm} \times 101.3 \text{ J/L} \cdot \text{ atm} = -17.76 \text{ J} \]

Key concepts

Gas Expansion

In thermodynamics, gas expansion refers to the process where a gas increases in volume. This change in volume often occurs due to an increase in temperature or a decrease in pressure. When a gas expands, it does so because the particles within it have additional energy that causes them to move further apart. Gas expansion can happen under various conditions:

• Isothermal expansion, where the temperature remains constant.
• Adiabatic expansion, where no heat is exchanged with the surroundings.
• Isobaric expansion, where the pressure remains constant.

In our exercise, the gas expands at constant temperature and influences the work done during this expansion based on surrounding pressures.

Work Done by Gas

When a gas expands, it can do work on its surroundings. This work is a result of the force the gas exerts over a distance as it pushes against the surrounding pressure. The work done by the gas is a key concept in thermodynamics and can be calculated using different conditions. Work can be positive or negative depending on whether the gas is expanding or compressing:

• Work is done by the gas when it expands against an external pressure. This is often considered negative in sign convention because energy is leaving the system.
• Conversely, work done on the gas (compression) is positive, meaning energy is entering the system.

In our problem statement, work is calculated for different scenarios where the gas expands against varying external pressures.

Pressure-Volume Work

Pressure-volume work is a type of work that occurs when the volume of a system changes while a pressure is exerted. It is essential to many chemical and physical processes that involve gases. The formula to calculate pressure-volume work is:\[ W = -P \Delta V \]where:

• \( W \) is the work done,
• \( P \) is the external pressure,
• \( \Delta V \) is the change in volume (final volume minus initial volume).

This formula shows that the work done by a gas is dependent on both the change in volume and the pressure against which it is expanding.

Constant Pressure

In cases of constant pressure, the pressure remains unchanged during the entire process of gas expansion or compression. This scenario is referred to as isobaric conditions in thermodynamics.When a gas expands at constant pressure, the pressure-volume work simplifies to:\[ W = -P (V_f - V_i) \]where:

• \( P \) is the constant external pressure,
• \( V_f \) and \( V_i \) are the final and initial volumes respectively.

The negative sign in the formula indicates that as the gas does work on the surroundings, energy is leaving the system. Constant pressure scenarios are common in open systems where the gas can expand freely into the environment, like in our example with the pressures of 1.5 atm and 2.8 atm.