9.4 The First Law of Thermodynamics

Internal Energy of a System

Internal energy represents the total energy contained within a thermodynamic system at the microscopic level. This energy is the sum of all the energies of the particles that make up the system.

The internal energy of a system consists of:

• Kinetic energy of the particles (atoms and molecules) moving within the system 🏃‍♂️
• Potential energy associated with the interactions and configurations between these particles

For ideal gases, internal energy has specific characteristics:

• Atoms in an ideal gas move independently with no attractive or repulsive forces between them
• The internal structure of individual atoms is not considered in calculations
• As a result, ideal gases have no internal potential energy component
• The internal energy of an ideal monatomic gas depends only on temperature and is calculated as: $U=\frac{3}{2} n R T=\frac{3}{2} N k_{B} T$

Where:

• $n$ is the number of moles
• $R$ is the universal gas constant
• $T$ is the absolute temperature
• $N$ is the number of atoms
• $k_B$ is the Boltzmann constant

When a system's internal energy changes, these changes can affect the internal structure and behavior of the system without necessarily changing the motion of the system's center of mass. This distinction is important for understanding how energy transformations occur at the microscopic level.

Thermodynamic Processes

Thermodynamic processes describe how systems change from one state to another. The first law of thermodynamics provides a framework for understanding these processes by restating the principle of energy conservation in thermal systems. 🔥❄️

For an isolated system (one that doesn't exchange energy or matter with its surroundings):

• The total energy remains constant over time
• Energy can be transformed from one form to another, but cannot be created or destroyed

For a closed system (one that exchanges energy but not matter with surroundings):

• The change in internal energy (ΔU) equals the sum of heat transferred to the system (Q) and work done on the system (W)
• This relationship is expressed mathematically as: ΔU = Q + W
• Work done on a system by external pressure that changes the volume is calculated as: $W=-P \Delta V$
• The negative sign indicates that when a system expands (ΔV is positive), work is done by the system (W is negative)

PV diagrams (pressure-volume graphs) are valuable tools for visualizing and analyzing thermodynamic processes:

• Each point on a PV diagram represents a specific state of the system
• Paths between points represent different thermodynamic processes
• Isotherms are curves on PV diagrams where temperature remains constant
• The area under a curve on a PV diagram equals the work done during the process

Several special cases of thermodynamic processes are particularly important:

• Isovolumetric (constant volume): ΔV = 0, so W = 0, and ΔU = Q
• Isothermal (constant temperature): For ideal gases, ΔU = 0, so Q = -W
• Isobaric (constant pressure): W = -PΔV, and ΔU = Q + W
• Adiabatic (no heat transfer): Q = 0, so ΔU = W

Understanding these processes helps analyze real-world systems from engines to atmospheric phenomena.

Practice Problem 1: Internal Energy Change

Solution

First, we need to apply the first law of thermodynamics to find the change in internal energy:

ΔU = Q + W

Where Q is the heat absorbed by the gas (positive) and W is the work done on the gas (negative when the gas expands).

Given:

• Q = 2500 J (heat absorbed)
• W = -1500 J (work done by the gas, so negative work done on the gas)
• n = 2.0 moles

ΔU = 2500 J + (-1500 J) = 1000 J

For a monatomic ideal gas, the internal energy is related to temperature by: $U = \frac{3}{2}nRT$

Therefore: $ΔU = \frac{3}{2}nR(T_f - T_i)$

Solving for the final temperature: $T_f = T_i + \frac{2ΔU}{3nR}$

$T_f = 300 \text{ K} + \frac{2 \times 1000 \text{ J}}{3 \times 2.0 \text{ mol} \times 8.31 \text{ \text{J/(mol}\cdot\text{K})}}$

$T_f = 300 \text{ K} + \frac{2000 \text{ J}}{49.86 \text{ J/K}}$

$T_f = 300 \text{ K} + 40.1 \text{ K}$

$T_f = 340.1 \text{ K}$

Practice Problem 2: Work in a Thermodynamic Process

Solution

For a process occurring against a constant external pressure, the work done by the gas can be calculated using:

$W_{by} = P_{ext}(V_f - V_i)$

Note that this is the work done by the gas, which is the negative of the work done on the gas in the first law of thermodynamics.

Given:

• Initial volume, $V_i = 2.0 \text{ L} = 2.0 \times 10^{-3} \text{ m}^3$
• Final volume, $V_f = 5.0 \text{ L} = 5.0 \times 10^{-3} \text{ m}^3$
• External pressure, $P_{ext} = 1.0 \times 10^5 \text{ Pa}$

$W_{by} = (1.0 \times 10^5 \text{ Pa})[(5.0 \times 10^{-3} \text{ m}^3) - (2.0 \times 10^{-3} \text{ m}^3)]$

$W_{by} = (1.0 \times 10^5 \text{ Pa})(3.0 \times 10^{-3} \text{ m}^3)$

$W_{by} = 300 \text{ J}$

Therefore, the gas does 300 J of work during the expansion. In terms of the first law of thermodynamics, the work done on the gas would be W = -300 J.