Question

A sample of nitrogen gas expands in volume from 1.6 to $5.4\mathrm{L}$ at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum, (b) against a constant pressure of 0.80 atm, and (c) against a constant pressure of 3.7 atm. (See Equation 5.4. $(1\mathrm{L}\cdot \mathrm{atm}=101.3\mathrm{J})$.

Short answer

(a) 0 J, (b) -308 J, (c) -1424.67 J.

Step-by-step solution

Understand the Problem

We need to calculate the work done when nitrogen gas expands from 1.6 L to 5.4 L under different conditions. There are three scenarios: expansion against a vacuum, a constant pressure of 0.80 atm, and a constant pressure of 3.7 atm.

Calculate Work Done Against a Vacuum

In a vacuum, there is no opposing force, so no work is done. Thus, the work done $W_a$ can be expressed as $W_a=0\text{J}$.

Use the Formula for Work Against Constant Pressure

The work done by the gas expanding against constant pressure is calculated using the formula $W=-P_{ext}\mathrm{\Delta }V$ where $P_{ext}$ is the external pressure and $\mathrm{\Delta }V=V_f-V_i$ is the change in volume.

Calculate Work for 0.80 atm

The change in volume $\mathrm{\Delta }V=5.4\text{L}-1.6\text{L}=3.8\text{L}$. Now, using $P_{ext}=0.80\text{atm}\text{convert work to joules,}$ $W=-(0.80)\times (3.8\text{L})\times (101.3\text{J/L}\cdot \text{atm})=-308\text{J}$.

Calculate Work for 3.7 atm

For this scenario, $P_{ext}=3.7\text{atm}$, so $W=-(3.7)\times (3.8\text{L})\times (101.3\text{J/L}\cdot \text{atm})=-1424.67\text{J}$. Round off if necessary.

Key concepts

Nitrogen Gas

Nitrogen gas is a diatomic molecule, meaning it is composed of two nitrogen atoms, and is represented by the symbol $N_2$. It is a colorless, odorless, and tasteless gas that makes up about 78% of Earth's atmosphere.
In thermodynamics, nitrogen gas is often used to exemplify the ideal gas behavior due to its stability and abundance. When performing calculations involving gases, it is important to consider the type of gas, as it influences properties such as compressibility and thermal conductivity.
In this context of the exercise, nitrogen gas undergoes an expansion process under constant temperature conditions, demonstrating how it behaves under different pressure scenarios.

Constant Pressure

When a gas is under constant pressure, it means that the force exerted by the gas on the walls of its container does not change during the process. This is significant in thermodynamic processes as it leads to specific relationships between volume and temperature changes.
Constant pressure scenarios are common in many practical systems where external pressure remains the same, such as atmospheric pressure or in a pressurized container. In this exercise, nitrogen gas expands against different constant pressures, demonstrating how varying pressures affect the work done by the gas.

• At 0.80 atm: The gas does less work as opposed to higher pressures.
• At 3.7 atm: More work is done since the pressure opposes the expansion.

To calculate work when pressure is constant, the formula used is $W=-P_{ext}\mathrm{\Delta }V$, where $P_{ext}$ is the external constant pressure.

Volume Expansion

Volume expansion refers to the increase in volume that occurs when a gas expands. It is a key factor in determining the work done by or on the gas.
The change in volume, $\mathrm{\Delta }V$, is calculated as the final volume minus the initial volume. In this exercise, the volume changes from $1.6\text{L}$ to $5.4\text{L}$, resulting in a volume change of $3.8\text{L}$.
This volume change is significant because it directly affects the amount of work done during expansion, impacting how much energy is transferred in or out of the system depending on the opposing pressure. Volume expansion calculations are crucial in understanding energy conversions in thermodynamic processes.

Joules Conversion

Joules is the unit of work or energy in the International System of Units (SI). When dealing with gases, especially in thermodynamics, we often need to convert between units such as liters-atmospheres to joules to understand the energy transformations.
The conversion factor used here is $1\text{L}\cdot \text{atm}=101.3\text{J}$. This means that for every liter atmosphere of work done by the gas, it is equivalent to 101.3 joules.
Using this conversion is essential in thermodynamics to ensure uniformity and accuracy when calculating work or energy in different units. For instance, in this exercise, converting the work done during gas expansion from atm to joules helps in understanding the energy scale involved.