1st Law of Thermodynamics

The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed but is transformable from one form to another. This is a cardinal tenet taken from classical thermodynamics and applies to every physical process. It introduces the basic concept that an independent system contains the same amount of total energy at whatever time it is considered, irrespective of the changes taking place within the system. Mathematically, this First Law is represented as ΔU=Q−W, where ΔU is the change in internal energy of the system, Q is the heat added to the system, and W is the work done by the system. This relationship emphasizes the interplay between heat, work, and internal energy.

First Law or Law of Conservation of Energy

It was introduced by Helmholtz and according to it "Energy can neither be created nor destroyed but can be converted from one form to another or the total energy of the universe is constant",

It can also be written as:

• The energy of an isolated system must remain constant, although it may be transformed from one form to another.

• Energy in one form, if it disappears will make its appearance in an exactly equivalent in another form.

• When work is transformed into heat or heat into work, the quantity of work is mechanically equivalent to the quantity of heat.

• It is never possible to construct a perpetual motion machine that could produce work without consuming any energy.

Thus if heat is supplied to a system it is never lost but it is partly converted into internal energy and partly in doing work in the system that is,

Heat supplied = Work done by the system + Increase in internal energy

So increase in internal energy = Heat supplied - work done by the system

ie. $\Delta E=q+w \quad[\because$ work done by the system is -w$]$

Mathematical Formulation of the First Law

If a system absorbs the 'q' amount of heat and its state changes from X to Y this heat is used up.

(i) On increasing the internal energy of the system

$\Delta \mathrm{E}=\mathrm{E}_{\mathrm{Y}}-\mathrm{E}_{\mathrm{X}}$

(ii) In order to do some external work (W) on the surroundings by the system.

From the first law, we get the relation

$
\begin{aligned}
& \Delta \mathrm{E}=\mathrm{Q}-\mathrm{W} \text { (that is, work done by the system }=\mathrm{W} \text { ) } \\
& d \mathrm{E}=d \mathrm{Q}-d \mathrm{~W} \text { or } d \mathrm{E}=d \mathrm{Q}-\mathrm{P} d \mathrm{~V})
\end{aligned}
$

Work done by the system or in expansion

OR

$
\begin{aligned}
& \Delta \mathrm{E}=\mathrm{Q}+\mathrm{W} \text { (that is, work done by the system }=\mathrm{W} \text { ) } \\
& d \mathrm{E}=d \mathrm{Q}+d \mathrm{~W} \text { or } d \mathrm{E}=d \mathrm{Q}+\mathrm{P} d \mathrm{~V})
\end{aligned}
$

Work done by the system or in compression

For the take of simplicity, remember the formula $\Delta \mathrm{E}=\mathrm{Q}+\mathrm{W}$ and when the work is done by the system, work is negative, and when the work is done on the system.

Recommended topic video on ( 1st law of Thermodynamics)

Some Solved Examples

Example 1: A heat engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_1$ . Work done by the engine is $J\left(Q_1+Q_2\right)$ This data

1)violates 1^st law of thermodynamics

2)violates 1^st law of thermodynamics if $Q_1$ is -ve

3)violates 1^st law of thermodynamics if $Q_2$ is -ve

4) (correct) does not violate 1^st law of thermodynamics

Solution

As we learned in the concept:

The energy of the universe is always conserved or the total energy of an isolated system is always conserved

$\Delta E=q+W$

$\Delta E$ Internal Energy

q= Heat

w= work

This is by the 1st law of thermodynamics

Hence, the answer is the option (4).

Example 2: Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)

1)Isochoric process: $\Delta U=q$

2) (Adiabatic process : $\Delta U=-w$

3)Isothermal process: $q=-w$

4)Cyclic process: $q=-w$

Solution

In the Adiabatic process:-

$q=0$

According to the first law of thermodynamics

$\Delta E=q+w$

So, $\Delta E=w$

Hence, the answer is the option (2).

Example 3: Which of the following will reduce the energy of the system?

1)Q=10J

2)$\Delta$V=5J

3)W=+10J

4) W=-10J

Solution

Any type of energy (Heat, Work) given to the system is positive and any type of energy taken out from the system

$W=+10 j$ i.e. 10 J work done on the system

$q=-10 j$ i.e. 10 J released by the system

W is -ve means work is being done by the system and hence its energy will decrease.

Hence, the answer is the Option (4).

Example 4: A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :

Q 1)10 J of the work will be done by the gas.

2) 6 J of the work will be done by the gas.

3) 10 J of the work will be done by the surrounding on gas.

4) 6 J of the work will be done by the surrounding on gas.

Solution

Given,

A gas undergoes change from state A to state B:

$\begin{aligned} & A \longrightarrow B \\ & \mathrm{Q}=5 \mathrm{~J} \\ & \mathrm{~W}=8 \mathrm{~J} \\ & \Delta \mathrm{U}_{\mathrm{AB}}=\mathrm{Q}+\mathrm{W}=5+(-8)=-3 \mathrm{~J}\end{aligned}$

Now, gas is brought back to A from state B:

$\begin{aligned} & B \longrightarrow A \\ & \mathrm{Q}=-3 \mathrm{~J} \\ & \Delta \mathrm{U}_{\mathrm{BA}}=3 \mathrm{~J} \\ & \Delta U_{B A}=-3+W \\ & 3+3=W \\ & W=6 J\end{aligned}$

(work is done on the system)

or

As work done has a positive sign, work is done by the surrounding on the gas.

Hence, the answer is an option (4).

Example 5: A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37⁰ C. As it does so ,it absorbs 208 J of heat. The values of q and w for the process will be :

(R= 8.314 J/mol K) (In 7.5 = 2.01)

1)$q=+208 J, w=+208 J$

2) $q=+208 J, w=-208 J$

3)$q=-208 J, w=-208 J$

4)$q=-208 J, w=+208 J$

Solution

The process is the isothermal reversible expansion

Hence dT = 0

$\Delta$U = 0

Now, work is given by the formula

$\mathrm{W}=-\mathrm{n} \times \mathrm{R} \times \mathrm{T} \times \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)$

Where, R= 8.314 J/mol K, n = 0.04 moles, T = 37⁰C = 310 K

$\mathrm{W}=-0.04 \times 8.314 \times 310 \times \ln \left(\frac{375}{50}\right)=-208 \mathrm{~J}$

Now, using 1st law of thermodynamics

$\mathrm{Q}+\mathrm{w}=\Delta \mathrm{U}$

$\therefore Q=-w=208 \mathrm{~J}$

Hence, the answer is the option (2).

Summary

The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed; it can only be transformed from one form to another. In any process, the total amount of energy in an isolated system remains constant. This means that the energy you put into a system, whether as heat or work, will either increase the system's energy or be converted to another form of energy. This law is fundamental to understanding how energy works in physical and chemical processes and is crucial for fields like physics, chemistry, and engineering.