**First law of thermodynamics**

work, w, is

where

F(x) = force as a function of x

x

_{i}& x_{f}= initial and final position, respectivelyθ = angle between F & x ; note: cos 0° = 1, while cos 180° = - 1

use cos θ, since work =

dot productof F & dxwork 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 & F_{wall} = - F_{person}, i.e. same magnitude, but opposite direction (to simplify the analysis, assume there is no friction. if there is friction, then F_{wall} + F_{person} + F_{friction} = 0)

to illustrate the concept of work in the context of the first law of thermodynamics, let's examine a 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 { x

_{f}- x_{i}} cos θ

**case 1**: reaction increases net number of gas molecules, e.g. 2 C_{2}H_{6} + 7 O_{2} --> 4 CO_{2} + 6 H_{2}O, which would cause an *increase* in the volume of the system, i.e. x_{f} - x_{i} > 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 = workby the system(on the surrounding), so use F_{system}dx vector is pointing up, while F

_{system}vector is pointing up, soθ = 0°, so

cos θ > 0 & x

_{f}- x_{i}> 0, sow > 0

^{ 1}∴ Δ U < 0.

case 1b. Δ U = q + w, where w = work

by the surrounding(on the system), so use F_{surrounding}dx vector is pointing up, while F

_{surrounding}vector is pointing down (apply Newton's third law), soθ = 180°, so

cos θ < 0 & x

_{f}- x_{i}> 0, sow < 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. x_{f} - x_{i} < 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 = workby the system(on the surrounding), so use F_{system}dx vector is pointing down, while F

_{system}vector is pointing down, soθ = 0°, so

cos θ > 0 & x

_{f}- x_{i}< 0, sow < 0

∴ Δ U > 0.

case 2b. Δ U = q + w, where w = work

by the surrounding(on the system), so use F_{surrounding}dx vector is pointing down, while F

_{surrounding}vector is pointing up (apply Newton's third law), soθ = 180° , so

cos θ < 0 & x

_{f}- x_{i}< 0, sow > 0

∴ Δ U > 0.

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^{ 1} somewhat more rigorous treatment.