First Law of Thermodynamics Notes | Class 12 | NEB | Nepali Educate

First Law of Thermodynamics

Syllabus: Content to read

1. Thermodynamic systems
2. Work done during volume change
3. Heat and work; Internal energy and First law of thermodynamics
4. Thermodynamic processes: Adiabatic, isochoric, isothermal and isobaric
5. Heat capacities of an ideal gas at constant pressure and volume and relation between them
6. Isothermal and Adiabatic processes for an ideal gas

Complete Note:

Defination:

It states that “when a certain amount of heat energy is supplied to a system, some part of it is used to perform external work and rest of heat is used to increase the internal energy of the system.”

Thermodynamic system and thermodynamic variables:

A definite part of matter separated from the whole of its amount and in which our attention is focused is called the thermodynamics system. A system that can exchange both energy and matter with its surroundings is called an open system.
The variable that is used to calculate a thermodynamic system completely is called the thermodynamic variable.

The molar specific heat capacity of a gas at constant pressure is always greater than that at constant volume:

At constant volume, the amount of heat energy supplied to a gas is used only to increase the temperature of the gas whereas, heat supplied at constant pressure is used to increase the temperature of the system and to perform external work. So, the amount of heat required to raise the temperature of the gas at constant pressure is always greater than the amount of heat required to raise the temperature of the same gas at the same amount at constant volume. Hence, the molar-specific heat capacity of a gas at constant pressure is always greater than that at constant volume.

Work done on the system and work done by the system:

If a system exerts a force on its surroundings and it is displaced by some amount, the system does external work or work done by the system.
When one part of a system does work on another part, it is said to have done internal work or work done on the system.

The internal energy of a real gas and ideal gas:

According to the kinetic theory of gases, the molecules of an ideal gas do not exert intermolecular forces. Therefore, the molecules of an ideal gas do not possess potential energy. As a result, the internal energy of an ideal gas is wholly kinetic and is a function of temperature only. The internal energy of a real gas, however, is partly made of kinetic energy and partly of intermolecular potential energy, and hence it is a function of temperature and volume.

Isothermal and adiabatic process:

The process in which the pressure and volume of a system change but the temperature remains constant is called the isothermal process.
The process in which all three parameters, pressure, volume, and temperature change but no heat enters the system and leaves the system is called the adiabatic process.

Differentiate between reversible and irreversible processes are:

In thermodynamics, processes are classified into two main categories: reversible and irreversible processes. A reversible process is one that can be retraced in the reverse order, while an irreversible process cannot be retraced in the opposite order by simply reversing it. These processes have distinct characteristics that set them apart. Below is a comparison table highlighting the key differences between reversible and irreversible processes.

Reversible process	Irreversible process
A reversible process can be retraced in the reverse order.	An irreversible process cannot be retraced in the opposite order by reversing.
Transfer of heat between two bodies at the same temperature.	Heating of electrical resistance.
No generation of entropy.	Generation of entropy.
Occurs slowly.	Occurs rapidly.
Entropy change is reversible.	Entropy change is irreversible.
Can be represented by a straight line on a T-S diagram.	Cannot be represented by a straight line on a T-S diagram.

Cyclic process:

If the state of a system at the end of a process is the same as the state of the system at the beginning, the process is called a cyclic process.
An example is the Carnot cycle, where an engine operates between two temperature reservoirs, absorbing heat, converting some to work, and rejecting the rest. The engine then repeats a series of reversible processes to return to its initial state. Another example is the Brayton cycle used in gas turbine engines, where air is compressed, heated, expanded to produce work, and then cooled and compressed again to complete the cycle.

The relation between the two specific heats of gas:

Consider n mole of an ideal gas is enclosed in a cylinder fitted with movable and functionless piston.
Let, dθ be the amount of heat supplied to the gas at constant volume. So, that its temperature rises by dT. Then,
dθ = nCvdT……1
Again, from first law of thermodynamics, we have,

dθ = du + dw

Since,

dw = Pdv

∴ dQ = dv + Pdv

At constant volume, dv = 0

So, dQ = du……..2

From 1 and 2

du = n${C_v}$dT………3

Let, the gas is heated at constant pressure so that its temperature rises by the same amount dT.

Then,

dQ’ = n${C_p}$dT……..4

From the first law of thermodynamics, we have

dQ’ = n${C_p}$dT

Or, n${C_p}$dT = du + pdv……5

From 3 and 5 we have,

nCpdt = n${C_v}$dT + pdv……6

Again from n mole of ideal gas we have

Pv = nRT

Differentiating both sides concerning T at constant pressure

Pdv = nRdT

From equations 6 and 7 we have

nCpdT = nCvdT =nRdT

Or, (${C_p}$ –${C_v}$) ndT =R ndT

Or, [${C_p}$ – ${C_v}$ = R]

This is the relation between the two specific heats of gas.

Deduce the expression of work done during the adiabatic process:

The process in which all three parameters, pressure, volume and temperature changes but no heat enters in to the system and leave out from the system is called adiabatic process.
Consider one mole of ideal gas is enclosed in a cylinder having non-conducting walls and fitted with moveable and frictionless piston.
Let${\rm{\: }}{{\rm{P}}_1}{\rm{\: }}{{\rm{V}}_1}{\rm{\: and\: }}{{\rm{T}}_1}$ are the initial state of the gas in the cylinder. Let, the gas expand from its initial volume ${V_1}$ to final volume ${V_2}$
Then, the work done by the is given by,
W = $\mathop \smallint \limits_{{{\rm{V}}_1}}^{{{\rm{V}}_2}} {\rm{Pdv}} \ldots \ldots \ldots \ldots \ldots 1$
For the adiabatic process, ${\rm{P}}{{\rm{V}}^{\rm{w}}} = {\rm{constant\: }}\left( {\rm{K}} \right)$
${\rm{or}},{\rm{\: P}} = \frac{{\rm{k}}}{{{{\rm{V}}^{\rm{w}}}}}$
Now, substituting the value of P in equation 1 we get
W = $\mathop \smallint \limits_{{{\rm{V}}_1}}^{{{\rm{V}}_2}} \frac{{\rm{K}}}{{{{\rm{V}}^{\rm{w}}}}}{\rm{dv}}$
$\frac{{\rm{K}}}{{1 - {\rm{w}}}}\left[ {{\rm{V}}_2^{1 - {\rm{w}}} - {\rm{V}}_1^{1 - {\rm{w}}}} \right]$
$\frac{{\rm{K}}}{{1 - {\rm{w}}}}\left[ {{\rm{KV}}_2^{1 - {\rm{w}}} - {\rm{KV}}_1^{1 - {\rm{w}}}} \right] \ldots \ldots \ldots .2$
Since, we have,
${{\rm{P}}_1}{\rm{\: V}}_1^{\rm{w}} = {\rm{\: }}{{\rm{P}}_2}{\rm{\: V}}_2^{\rm{w}} = {\rm{K}}\left( {{\rm{constant}}} \right)$
Here from 2
${\rm{W}} = \frac{1}{{1 - {\rm{w}}}}\left( {{{\rm{P}}_1}{\rm{\: }}{{\rm{V}}_1} - {\rm{\: }}{{\rm{P}}_2}{\rm{\: }}{{\rm{V}}_2}} \right) \ldots \ldots \ldots .3$
For one mole of ideal gas,
Pv +RT
${\rm{W}} = \frac{{\rm{R}}}{{1 - {\rm{w}}}}\left( {{{\rm{T}}_1}{\rm{\: }} - {\rm{\: }}{{\rm{T}}_2}{\rm{\: }}} \right) \ldots \ldots \ldots .4$
This equation 3 and 4 gives the work done is the work done during adiabatic process.

In adiabatic process PV^y = constant;

The process in which all three parameters, pressure, volume and temperature changes but no heat enters in to the system and leave out from the system is called adiabatic process.
Consider one mole of ideal gas is enclosed in a cylinder having non-conducting walls and fitted with moveable and frictionless piston.
Let${\rm{\: }}{{\rm{P}}_1}{\rm{\: }}{{\rm{V}}_1}{\rm{\: and\: }}{{\rm{T}}_1}$ are the initial state of the gas in the cylinder.
Let, the gas is compressed suddenly so that, its temperature is raised by dT.
Hence change in internal energy,
Du = 1 * Cv dT
Where, Cv is the molar heat capacity of the gas.
Now, from first law of thermodynamics, we have,
dθ = dU + d w
For adiabatic process, dQ = 0
Hence,
dU = -PdV…….1
Again, dW = PdV……2
So,
dU = 1 * ${C_v}$dT
Now, substituting the value of dU from equation 1 we get,
CvdT = -Pdv…….3
Now, for 1 mole of an ideal gas, PV =RT
Differentiating both sides with respect to T, we get
$\frac{{{\rm{dV}}}}{{{\rm{dT}}}} + {\rm{V}}\frac{{{\rm{dP}}}}{{{\rm{dT}}}} = {\rm{R}}\frac{{{\rm{dT}}}}{{{\rm{dT}}}}$
${\rm{or}},{\rm{\: dT}} = \frac{{{\rm{PdV}} + {\rm{VdP}}}}{{\rm{R}}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots 4$
Substituting the value of dT in equation 3 we get,
${\rm{Cv\: }}\left[ {\frac{{{\rm{PdV}} + {\rm{VdP}}}}{{\rm{R}}}} \right] = - {\rm{PdV}}$
${\rm{or}},{\rm{\: }}\left( {{\rm{Cv}} + {\rm{R}}} \right){\rm{Pdv}} + {\rm{Cv\: VdP}} = 0 \ldots \ldots \ldots .5$
Again, we have,
${C_p}$ –${C_v}$ =R
∴ ${C_p}$=${C_v}$ +R
From 5,
${C_p}$.Pdv +${C_c}$V${P_p}$ =0
Dividing both sides by Cv Pv, we get
$\gamma \frac{{{\rm{dV}}}}{{\rm{V}}} + \frac{{{\rm{dP}}}}{{\rm{p}}} = {\rm{o}}$.....6
Where γ = $\frac{{{\rm{Cp}}}}{{{\rm{CV}}}}{\rm{\: is\: }}$ the ratio of molar heat capacity of a gas at constant volume.
Now integrating both side of equation 6 we get
$\gamma {\log _{\rm{e}}}{\rm{V}} + {\log _{\rm{e}}}{\rm{P}} = {\rm{constant}}$
Or, ${\rm{P}}{{\rm{V}}^\gamma } = {\rm{constant}}$ which is required expression.

The work done by gas during expansion is numerically equal to the area under the P.V. diagram:

Work done by the gas during expansion:
When the gas expands, the piston moves out through a small distance dx and work done by the force dW is given by
dW = Fdx = PAdx
Since Adx = dV, a small increase in the volume of the gas, the work done by the gas during the expansion.
dW = PdV…(i)
When the volume of the gas changes from ${V_1}$ to ${V_2}$, the total work done W is obtained by integrating the equation (i) within the limits ${V_1}$ to ${V_2}$.
W=∫dW=\int\limits_{{V_1}}^{{V_2}} {Pdv} ........……(ii)
When the final volume ${V_2}$ is greater than initial volume ${V_1}$, then the change in volume ${V_2}$- ${V_1}$ is positive. Hence during the expansion of the gas work done by a system is taken as positive.
When the gas is compressed, the final volume ${V_2}$ is less than the initial volume V1then the change in volume ${V_2}$- ${V_1}$ is negative. Hence during the compression of the gas work done by a system is taken as negative.

The limitations of the first law of thermodynamics:

The limitations of the first law of thermodynamics are:

1. It does not indicate the direction of heat transfer.
2. It does not indicate heat energy developed in the target cannot be converted back into mechanical energy of the bullet enabling it to fly back.
3. It does not give to what extent the mechanical energy is obtained from the heat energy.

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