The First Law: Heat, Work, and Internal Energy
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You should be comfortable with the kinetic theory derivation of the ideal gas law and the basic calculus of work done by a variable force. Familiarity with the definitions of molar heat capacities \(C_V\) and \(C_p\) is assumed.
Why this matters
The First Law is the bookkeeping rule that prevents energy from disappearing in thermal processes.
While the previous lecture established how molecular collisions generate pressure and temperature, this lecture introduces the macroscopic energy balance. It defines internal energy as a state function and distinguishes between heat and work as the two modes of energy transfer. This framework is essential for analyzing engines, refrigerators, and phase transitions, providing the rigorous foundation upon which the Second Law and statistical mechanics are built.
The Energy Balance
Imagine a gas in a cylinder with a movable piston. If you push the piston in, you do work on the gas. If the gas expands and pushes the piston out, the gas does work on the surroundings. In both cases, energy is transferred. However, you can also heat the gas by placing it in a flame. Both work and heat change the internal energy of the gas, but they are distinct mechanisms. Work is organised energy transfer associated with macroscopic displacement, while heat is disorganized energy transfer driven by a temperature difference. The First Law states that the change in the system's internal energy is equal to the heat added to the system minus the work done by the system. It is a statement of energy conservation applied to thermodynamic systems.
Internal Energy and State Functions
Internal energy \(U\) is a state function, meaning its value depends only on the current state of the system (defined by variables like \(T, V, p\)) and not on the path taken to reach that state. For a monatomic ideal gas, kinetic theory tells us that the internal energy is entirely translational kinetic energy. Since the mean kinetic energy per molecule is \(\frac{3}{2}k_B T\), the total internal energy for \(N\) molecules is \(U = \frac{3}{2}N k_B T\). In terms of moles, \(U = \frac{3}{2}nRT\). This implies that for an ideal gas, \(U\) depends only on temperature. The differential change in internal energy is therefore \(dU = n C_V dT\), where \(C_V\) is the molar heat capacity at constant volume. For a monatomic ideal gas, \(C_V = \frac{3}{2}R\).
This relationship holds for any process involving an ideal gas, whether the volume is constant or not, because \(U\) is a state function. The path independence of \(\Delta U\) is a crucial distinction from work and heat, which are path-dependent.
Work in Quasi-Static Processes
Work is done by a system when it expands against an external pressure. For a quasi-static process (one that proceeds infinitely slowly so the system remains in equilibrium), the work done by the gas is given by the integral of pressure over the change in volume. If the gas expands from volume \(V_1\) to \(V_2\), the work done by the gas is:
Using the ideal gas law \(p = \frac{nRT}{V}\), we can evaluate this integral for specific processes. For an isothermal expansion (constant \(T\)), \(p\) varies as \(1/V\), so \(W = nRT \ln(V_2/V_1)\). For an isobaric expansion (constant \(p\)), \(W = p(V_2 - V_1)\). Note the sign convention: work done by the system is positive. This is consistent with the First Law form \(dU = dQ - dW\).
The First Law and Heat Capacities
The First Law of Thermodynamics is written as \(dU = dQ - dW\), where \(dQ\) is the infinitesimal heat added to the system and \(dW = p \, dV\) is the infinitesimal work done by the system. Substituting the expression for \(dU\) and \(dW\), we get:
Rearranging for heat, \(dQ = n C_V \, dT + p \, dV\). This equation allows us to calculate the heat exchanged for any quasi-static process. For an isochoric process (\(dV = 0\)), \(dQ = n C_V \, dT\). For an isobaric process, we use the ideal gas law to relate \(dT\) and \(dV\). Differentiating \(pV = nRT\) at constant \(p\) gives \(p \, dV = nR \, dT\). Substituting this into the heat equation:
By definition, the molar heat capacity at constant pressure is \(C_p\), so \(dQ = n C_p \, dT\). Comparing the two expressions, we find the fundamental relation for ideal gases:
This result is a direct consequence of the First Law and the ideal gas equation of state. It shows that \(C_p\) is always greater than \(C_V\) by the gas constant \(R\), because at constant pressure, the gas must do work as it expands, requiring more heat input to achieve the same temperature rise.
The derivations
The Ideal Gas Law from Kinetic Theory
For a dilute gas of N identical point molecules of mass m confined to a rigid container of volume V, averaging the momentum delivered to the walls by elastic molecular collisions yields the equation of state PV = NkBT, provided one identifies the absolute temperature through ½m⟨v²⟩ = &frac32;kBT — i.e. the mean translational kinetic energy per molecule is &frac32;kBT.
The First Law of Thermodynamics and Work in Ideal-Gas Processes
For a closed system of \(n\) moles of an ideal gas, the change in internal energy \(U\) is a state function satisfying \(dU = n C_V\,dT\), and the first law of thermodynamics reads \(dU = dQ - dW\), where \(dQ\) is the heat added to the system and \(dW = p\,dV\) is the work done by the system during a quasi-static volume change. For a quasi-static path between equilibrium states \((p_1,V_1,T_1)\) and \((p_2,V_2,T_2)\), the work and heat depend on the path while \(\Delta U\) depends only on the end states; for the four standard processes (isochoric, isobaric, isothermal, adiabatic) these quantities reduce to closed-form expressions involving \(C_V\), \(C_p\), \(R\), and \(\gamma = C_p/C_V\).
Heat Conduction: Fourier's Law and the Heat Equation
In a homogeneous, isotropic solid of constant thermal conductivity \(k\), density \(\rho\), and specific heat capacity \(c\), the local heat flux density \(\vec{q}\) (power per unit area, \(\mathrm{W/m^2}\)) is proportional to the negative temperature gradient: \(\vec{q} = -k\,\vec{\nabla}T\). Combining this constitutive law with local energy conservation (first law, no mechanical work) yields the heat equation \(\frac{\partial T}{\partial t} = \alpha\,\nabla^{2}T\), where the thermal diffusivity is \(\alpha = k/(\rho c)\). In one dimension with steady state and no internal sources, the temperature profile is linear, and the heat current through a slab of area \(A\), thickness \(L\), and face temperatures \(T_{\text{hot}}\) and \(T_{\text{cold}}\) is \(H = kA(T_{\text{hot}}-T_{\text{cold}})/L\).
A common misconception is that heat is a property of the system, like internal energy. Heat is not a state function; it is energy in transit. You cannot say a system "contains" heat. Only the change in internal energy is well-defined for a state. Confusing heat with internal energy leads to errors in calculating efficiency and entropy changes. C
Worked example
Problem. Two moles of a monatomic ideal gas expand isothermally at \(T = 300 \, \mathrm{K}\) from \(V_1 = 10 \, \mathrm{L}\) to \(V_2 = 20 \, \mathrm{L}\). Calculate the work done by the gas and the heat absorbed.
- Identify the process: Isothermal expansion of an ideal gas. Since \(T\) is constant, \(\Delta U = 0\) because \(U\) depends only on \(T\).
- Apply the First Law: \(dU = dQ - dW \implies 0 = dQ - dW \implies dQ = dW\).
- Calculate work: For isothermal expansion, \(W = nRT \ln(V_2/V_1)\). Use \(n = 2 \, \mathrm{mol}\), \(R = 8.314 \, \mathrm{J/(mol \cdot K)}\), \(T = 300 \, \mathrm{K}\), \(V_1 = 0.010 \, \mathrm{m^3}\), \(V_2 = 0.020 \, \mathrm{m^3}\).
- Compute: \(W = 2 \times 8.314 \times 300 \times \ln(2) \approx 4988 \times 0.693 \approx 3457 \, \mathrm{J}\).
- Since \(dQ = dW\), \(Q = 3457 \, \mathrm{J}\).
Units check. \(nRT\) has units \(\mathrm{mol} \cdot \mathrm{J/(mol \cdot K)} \cdot \mathrm{K} = \mathrm{J}\). The logarithm is dimensionless. The result is in Joules, which is correct for energy.
Three quick tests
- Why is \(C_p > C_V\) for an ideal gas?
Answer
At constant pressure, the gas does work as it expands, so more heat is required to raise the temperature by the same amount compared to constant volume. - If an ideal gas undergoes an adiabatic expansion, what is \(dQ\)?
Answer
\(dQ = 0\). By the First Law, \(dU = -dW\), so the internal energy decreases as work is done by the gas. - Is work a state function?
Answer
No. Work depends on the path taken between two states, not just the initial and final states.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-149 · D-433 · D-434 · Threads: energy, matter, chance