First law of thermodynamics

work, w, is

work formula   

where

F(x) = force as a function of x

xi & xf = initial and final position, respectively

θ = angle between F & x ; note: cos 0° = 1, while cos 180° = - 1

use cos θ, since work = dot product of F & dx

work maybe viewed as the area under the graph of F(x) or to simplify the situation, assume there is a constant force, in which case, work is the product of force and displacement.

examples of work: (i) car engine, where the combustion of air & gas displaces the piston, (ii) muscle contraction, and (iii) missle / bullet's flight - all of these examples involve a chemical reaction.

Newton's third law: for any force on an object, there is an equal force exerted by the object, e.g. a person pushes on a wall - the person exerts a force on the wall, while the wall exerts a force on the person; the net force is zero, since neither the wall or person moves & Fwall = - Fperson, i.e. same magnitude, but opposite direction (to simplify the analysis, assume there is no friction. if there is friction, then Fwall + Fperson + Ffriction = 0)

to illustrate the concept of work in the context of the first law of thermodynamics, let's examine a car engine:

car engine

system = location of chemical reaction; below the piston.

surrounding = piston.

The first law of thermodynamics

Δ U = q - w

or

Δ U = q + w

which depends on the definition of w, work; i.e. the 1st law can be expressed either as a sum or difference - depending on the definition of work - as work on the system (by the surrounding) or work by the system (on the surrounding).

Δ U is the change in the internal energy of the system

q, heat, is the transfer of kinetic energy due to the collision of molecules, where the direction of energy transfer, heat, is from high to low temperature

w, work, is the transfer of energy other than by heat.

To simplify the analysis, assume it's an adiabatic process, i.e. q = 0 and F(x) = constant, so

Δ U = ± w = ± F { xf - xi} cos θ

case 1: reaction increases net number of gas molecules, e.g. 2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O, which would cause an increase in the volume of the system, i.e. xf - xi > 0 (and dx vector is pointing up) . . . Δ U < 0, i.e. the system's internal energy decreases as its volume increases.

case 1a. Δ U = q - w, where w = work by the system (on the surrounding), so use Fsystem

dx vector is pointing up, while Fsystem vector is pointing up, so

θ = 0°, so

cos θ > 0 & xf - xi > 0, so

w > 0 1

∴ Δ U < 0.

case 1b. Δ U = q + w, where w = work by the surrounding (on the system), so use Fsurrounding

dx vector is pointing up, while Fsurrounding vector is pointing down (apply Newton's third law), so

θ = 180°, so

cos θ < 0 & xf - xi > 0, so

w < 0

∴ Δ U < 0.

case 2: reaction decrease net number of gas molecules, which would cause a decrease in the volume of the system, i.e. xf - xi < 0 (and dx vector is pointing down) . . . . Δ U > 0, i.e. the system's internal energy increases as its volume decreases.

case 2a. Δ U = q - w, where w = work by the system (on the surrounding), so use Fsystem

dx vector is pointing down, while Fsystem vector is pointing down, so

θ = 0°, so

cos θ > 0 & xf - xi < 0, so

w < 0

∴ Δ U > 0.

case 2b. Δ U = q + w, where w = work by the surrounding (on the system), so use Fsurrounding

dx vector is pointing down, while Fsurrounding vector is pointing up (apply Newton's third law), so

θ = 180° , so

cos θ < 0 & xf - xi < 0, so

w > 0

∴ Δ U > 0.

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1 somewhat more rigorous treatment.

volume work