# first law of thermodynamics

First law of thermodynamics:

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 whole of its amount and in which our attention is focused is called thermodynamics system. A system in which can exchange both energy and matter with its surrounding is called open system.

The variable which is used to calculate a thermodynamic system completely is called thermodynamic variable.

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

At constant volume, the amount of heat energy supplied to a gas is used only to increases 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 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 same amount at constant volume. Hence, the molar specific heat capacity of a gas at constant pressure 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 kinetic theory of gases, the molecules of an ideal gas do not exert intermolecular forces. Therefore the molecules of an ideal gas do not posses potential energy. As a result, the internal energy of an ideal gas is wholly kinetic in nature 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 pressure and volume of a system changes but the temperature remains constant is called isothermal 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.

Differentiate between reversible and irreversible processes are;

 reversible process irreversible process 1. A reversible process is one which can be retraced in the reverse order.   2. Transfer of heat between two bodies at the same temperature. 1. An irreversible process is one which cannot be retraced in the opposite order by reversing.   2. Heating of an electrical resistance.

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.

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 + d w

Since,

dw = Pdv

∴ dQ = dv + Pdv

At constant volume, dv = 0

So, dQ = du……..2

From 1 and 2

Du = nCvdT………3

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

Then,

dQ’ = nCpdT……..4

From first law of thermodynamics, we have

dQ’ = nCpdT

Or, nCpdT = du + pdv……5

From 3 and 5 we have,

nCpdt = nCvdT + pdv……6

Again from n mole of ideal gas we have

Pv = nRT

Differentiating both sides with respect to T at constant pressure,

${\rm{P}}\frac{{{\rm{dv}}}}{{{\rm{dT}}}} = {\rm{nR}}\frac{{{\rm{dT}}}}{{{\rm{dT}}}}$

Or, Pdv = nRdT

From equation 6 and 7 we have

nCpdT = nCvdT =nRdT

Or, (Cp –Cv) ndT =R ndT

Or, [Cp – Cv = R]

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

Deduce the expression of work done during isothermal process;

The process in which pressure and volume of a system changes but the temperature remains constant is called isothermal process.

Consider n mole of ideal gas is enclosed in a cylinder having conducting walls and fitted with moveable and frictionless piston.

Let the gas expand from initial state ${\rm{A\: }}\left( {{{\rm{P}}_1}{\rm{\: }}{{\rm{V}}_1}} \right)$ to final state ${\rm{B\: }}\left( {{\rm{\: }}{{\rm{P}}_2}{\rm{\: }}{{\rm{V}}_2}} \right)$ very slowely such that the temperature of the system remains constant. Then, the work done by the gas in expanding from ${\rm{A\: }}\left( {{{\rm{P}}_1}{\rm{\: }}{{\rm{V}}_1}} \right)$ to ${\rm{B\: }}\left( {{\rm{\: }}{{\rm{P}}_2}{\rm{\: }}{{\rm{V}}_2}} \right)$ is given by,

W = $\mathop \smallint \limits_{{{\rm{V}}_1}}^{{{\rm{V}}_2}} {\rm{Pdv}} \ldots \ldots \ldots \ldots \ldots 1$

For n moles of gas,

PV = n R T

Or, P = $\frac{{{\rm{nRT}}}}{{\rm{V}}}$

Hence from 1,

W = $\mathop \smallint \limits_{{{\rm{V}}_1}}^{{{\rm{V}}_2}} \frac{{{\rm{nRT}}}}{{\rm{v}}}{\rm{\: dv\: }}$

Or, W = nRT $\mathop \smallint \limits_{{{\rm{V}}_1}}^{{{\rm{V}}_2}} \frac{{{\rm{dv}}}}{{\rm{v}}}{\rm{\: \: }}$

Or, ${\rm{w\: }} = {\rm{\: nRT}}[{\log _{\rm{e}}}{{\rm{V}}_2} - {\rm{Lo}}{{\rm{g}}_{\rm{e}}}{\rm{\: }}{{\rm{V}}_1}{\rm{\: }}]$

Or, ${\rm{w\: }} = {\rm{\: nRT}}{\log _{\rm{e}}}(\frac{{{{\rm{V}}_2}}}{{{{\rm{V}}_1}}})$

For isothermal process,

${{\rm{P}}_1}{\rm{\: }}{{\rm{V}}_1} = {{\rm{P}}_2}{{\rm{V}}_2}$

∴$\frac{{{{\rm{V}}_2}}}{{{{\rm{V}}_1}}} = \frac{{{{\rm{P}}_1}}}{{{{\rm{P}}_2}}}$

Hence work done in an isothermal process.

Deduce the expression of work done during 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 V1 to final volume V2.

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 PVy = 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 * CvdT

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,

Cp –Cv =R

∴ Cp=Cv +R

From 5,

Cp Pdv +Cv V Pp =0

Dividing both sides by Cv Pv, we get

$\gamma \frac{{{\rm{dV}}}}{{\rm{V}}} + \frac{{{\rm{dP}}}}{{\rm{p}}} = {\rm{o}}$

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 gas during expansion

Work Done by Gas during Expansion

When the gas expands, the piston moves out through a small distance dx

And work done by the force

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 V1 to V2, the total work done W is obtained by integrating the equation (i) within the limits V1 to V2.

W=∫dW=∫V2V1Pdv……(ii)

When the final volume V2 is greater than initial volume V1, then the change in volume V2- V1 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 V2 is less than the initial volume V1then the change in volume V2- V1 is negative. Hence during the compression of the gas work done by a system is taken as negative.

The limitations of first law of thermodynamics:

The limitations of 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 in to 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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