Question

Derive first law of thermodynamics mathematically

Answer 1

$\pink{ \sf \underline{ \underline{ \large{ \: First \: law \: of \: thermodynamics \: \: }}}}$

First law of thermodynamics states that energy can neither be created nor be destroyed

The mathmetical formulation of first law of thermodynamics : Δ U = q + w

Where ,

$\star$Δ U = change is internal energy

$\star$q = heat

$\star$w = work done

Now , we are going to derive this equation or formula

Suppose ,

The initial internal energy of the system = U1

if it gain or absorbs heat q its internal energy will become

U1 + q

If further work w is done on the system , the internal energy will further increases and become

U1 + q + w

Consider , this final energy as U2

Then ,

$\sf \implies U2 = U1 + q + w \\ \\ \sf \implies </p><p>U2 - U1 = q + w \\ \\ \sf \implies \red{\fbox{</p><p>Δ U = q + w}}$

Answer 2

First Law of Thermodynamics: It states that "energy can neither be created nor be destroyed".

Mathematical Expression: Q=W+dE

First, consider a system moving from the first state to the second state following the path 'a' and the same system return back to the second stage following the path 'b' and 'c'.

1. Consider the cyclic process, we can write 1-a-2-b-1 in the form,

$\int\limits_{1-a-2-b-1}{Q-W} \, = 0$  This can be split into 1-a-2 and 2-b-1

∴ $\int\limits_{1-a-2}{Q-W} + \int\limits_{2-b-1}{Q-W} =0$ → (equation 1)

The process 1-a-2-c-1 is also cyclic, it is written as:

$\int\limits_{1-a-2-c-1}{Q-W} \, = 0$ This can be split into 1-a-2 and 2-c-1

∴ $\int\limits_{1-a-2}{Q-W} + \int\limits_{2-c-1}{Q-W} =0$ → (equation 2)

On comparing equations 1 and 2 we get to know that the values obtained will not be equal to zero.

∴$\int\limits_{2-b-1}{Q-W} = \int\limits_{2-c-1}{Q-W} \neq 0$

By generalizing the above equation for the non-cyclic process we can write:

⇒Q-W=dE

⇒Q = W+dE (Expression for the first law of thermodynamics)

∴ The mathematical relationship is: Q=W+dE