The Timpson Lyceum seal

The Timpson Lyceum · University Physics I (2025)

First Law of Thermodynamics - Lecture 19 (University Physics I)

Overview

This lecture introduces the first law of thermodynamics, the broad statement of energy conservation that governs systems of gas (here, ideal gases almost exclusively). It defines the internal energy of a gas, shows how to compute the work done by a gas as the area under a curve on a pV diagram, and ties heat, work, and internal energy together through ΔU=Q−W\Delta U = Q - W. With that machinery in place, it walks through the four canonical processes a gas can undergo — isothermal, isobaric, isochoric (constant volume), and adiabatic — deriving the work and heat for each, and explores why work and heat are not state variables while internal energy is. It closes with the molar heat capacities CVC_V and CPC_P, their relation CP=CV+RC_P = C_V + R, the ratio γ=CP/CV\gamma = C_P/C_V, and the adiabatic relations between pressure, volume, and temperature.

Key Concepts

Internal energy (UU)

The first law of thermodynamics

Work done by a gas

State vs. path variables

The four processes (each computes work via W=∫p dVW=\int p\,dV)

Adiabatic relations

TVγ−1=constantpVγ=constant\begin{aligned} TV^{\gamma-1} &= \text{constant} \\ pV^{\gamma} &= \text{constant} \end{aligned}

Heat capacities of a gas

Worked Example

Vaporizing water (constant pressure). One gram (1 cm31\text{ cm}^3) of water at its boiling point is converted entirely to steam at atmospheric pressure p=1.013×105 Pap = 1.013\times10^5\text{ Pa}. Steam occupies 1671×1671\times the liquid volume, so ΔV=1670 cm3=1670×10−6 m3\Delta V = 1670\text{ cm}^3 = 1670\times10^{-6}\text{ m}^3. The heat of vaporization is Lv=2.256×106 J/kgL_v = 2.256\times10^6\text{ J/kg}.

Part A — work done by the expanding water (isobaric, so W=p ΔVW = p\,\Delta V):

W=p (V2−V1)W=(1.013×105)(1670×10−6)≈169 J\begin{aligned} W &= p\,(V_2 - V_1) \\ W &= (1.013\times10^{5})(1670\times10^{-6}) \approx 169\ \text{J} \end{aligned}

Part B — change in internal energy. The heat added is Q=mLvQ = mL_v, then apply the first law ΔU=Q−W\Delta U = Q - W:

ΔU=mLv−WΔU=(0.001)(2.256×106)−169≈2087 J\begin{aligned} \Delta U &= mL_v - W \\ \Delta U &= (0.001)(2.256\times10^{6}) - 169 \approx 2087\ \text{J} \end{aligned}

Almost all of the added heat becomes internal energy; only about 169 J goes into pushing back the atmosphere.

Summary