Heat and Work in Thermodynamics

The Nature of Work ($w$)

Unlike the random motion associated with heat, work is the transfer of energy through organized motion. In physics and chemistry, work occurs when a force acts upon an object to cause a displacement. In a chemical context, this most commonly appears as "Pressure-Volume work" ($P$-$V$ work), where a gas expands or contracts against an external pressure.

When a gas expands, it pushes against its surroundings, requiring energy to move the "piston" or the atmosphere. This energy comes from the system's internal reserves. Conversely, if the surroundings compress the gas, work is being done on the system, increasing its energy.

The mathematical expression for $P$-$V$ work is:

$$w = -P_{ext} \cdot \Delta V$$

Where:

• $w$ is the work done.
• $P_{ext}$ is the external pressure.
• $\Delta V$ is the change in volume ($V_{final} - V_{initial}$).

Comparing the Mechanisms

The fundamental distinction lies in how the energy is "packaged" during the transfer.

While heat moves through the "jittery" vibrations and translations of molecules, work moves through a coordinated push or pull that affects the entire system's boundaries simultaneously.

The First Law of Thermodynamics

To understand how heat and work interact, we look to the First Law of Thermodynamics, which is essentially the law of conservation of energy applied to thermal systems. The law states that the change in the internal energy ($\Delta U$) of a system is equal to the sum of the heat added to the system and the work done on the system.

The equation is expressed as:

$$\begin{aligned} \Delta U &= q + w \end{aligned}$$

In this convention:

1. If $q > 0$, heat is absorbed by the system (endothermic).
2. If $q < 0$, heat is released by the system (exothermic).
3. If $w > 0$, work is done on the system (e.g., compression).
4. If $w < 0$, work is done by the system (e.g., expansion).

Worked Example: Gas Expansion

Let's apply these concepts to a practical scenario. Imagine a gas in a cylinder with a movable piston. The gas is at a constant external pressure of $1.50\text{ atm}$. The gas expands from an initial volume of $2.00\text{ L}$ to a final volume of $5.00\text{ L}$. How much work is done by the gas?

Step 1: Identify the variables.

• $P_{ext} = 1.50\text{ atm}$
• $V_{initial} = 2.00\text{ L}$
• $V_{final} = 5.00\text{ L}$

Step 2: Calculate $\Delta V$.

$$\Delta V = 5.00\text{ L} - 2.00\text{ L} = 3.00\text{ L}$$

Step 3: Use the work formula.

$$\begin{aligned} w &= -P_{ext} \cdot \Delta V \\ w &= -(1.50\text{ atm}) \cdot (3.00\text{ L}) \\ w &= -4.50\text{ L}\cdot\text{atm} \end{aligned}$$

Step 4: Convert to Joules (optional but common). Since $1\text{ L}\cdot\text{atm} = 101.325\text{ J}$:

$$\begin{aligned} w &= -4.50 \cdot 101.325\text{ J} \\ w &\approx -455.96\text{ J} \end{aligned}$$

Interpretation: The negative sign indicates that the system did work on the surroundings. The energy required to push the piston out came from the internal energy of the gas.