Part III FIRST LAW OF THERMODYNAMICS

Distinct path names are given for each process: adiabatic (adiabat), isochoric (isochore), isobaric (isobar), and isothermal (isotherm).

The video begins by introducing the four key thermodynamic processes: adiabatic, isochoric, isobaric, and isothermal, focusing on their definitions and how they relate to the first law of thermodynamics, which states the change in internal energy (U) equals heat (Q) minus work (W).

An adiabatic process involves no heat transfer (Q=0). Consequently, the change in internal energy is equal to the negative of the work done (ΔU = -W). This means any change in internal energy is solely due to work done on or by the system.

In an isochoric process, the volume remains constant (ΔV=0), meaning no work is done (W=0). Therefore, the change in internal energy is equal to the heat added or removed (ΔU = Q). The heat can be calculated as n * Cv * ΔT, where Cv is the molar heat capacity at constant volume.

An isobaric process maintains constant pressure. In this case, work is done (W = PΔV), and the change in internal energy is given by ΔU = Q - W. Heat (Q) can be calculated using n * Cp * ΔT, where Cp is the molar heat capacity at constant pressure.

An isothermal process occurs at a constant temperature (ΔT=0), which means the change in internal energy is zero (ΔU=0). Therefore, the heat transferred into or out of the system is equal to the work done (Q = W).

The video illustrates these processes using Pressure-Volume (PV) diagrams. An isochoric line is vertical (constant volume), an isobaric line is horizontal (constant pressure), and isothermal curves are less steep than adiabatic curves, which are noted to be steeper because there is no heat added.

The molar heat capacities at constant volume (Cv) and constant pressure (Cp) are discussed. Key relationships are provided: Cp = Cv + R (where R is the gas constant) and gamma (γ) = Cp/Cv, which is the ratio of heat capacities. R is 8.314 J/mol·K.

A problem is presented: calculate the change in internal energy for 2500 moles of air cooled from 35°C to 26°C at constant pressure (1 atm), assuming air is an ideal gas with γ = 1.40. This is identified as an isobaric process.

The solution involves first calculating Cv using the relationship γ = 1 + R/Cv, which yields Cv = R / (γ-1). With the calculated Cv, the change in internal energy (ΔU) is then found using the formula ΔU = n * Cv * ΔT, resulting in ΔU = -4.60 × 10^5 Joules.

The video briefly mentions another formula for work in an adiabatic process: W = nCv(T1 - T2), or W = 1/(γ-1) * (P1V1 - P2V2), reinforcing that adiabatic curves are always steeper than isothermal curves on a PV diagram passing through the same point.