Topics

Experiments shown that there is a specific relationship between the state variables. ( \(P,V,T\) ) There is a very clear proportionality between the quantity PV and the quantity nT . If we designate the slope of the line in this graph as R , then we can write relationship as

\[PV = R \times (nT)\]

A graph of PV versus nT for an ideal gas.

It is customary to write the relationship in a slightly different form, namely

\[\begin{array}{r} PV = nRT\ \text{~}\text{(ideal-gas law)}\text{~}\#(i) \end{array}\]

equation (i) is ideal gas law.
The constant R , which determined experimentally as the slope of the graph in figure is called the universal gas constant. Its value in SI units, is

\[R = 8.31\text{ }J - mol\text{ }K\]

If the gas is initially in state i , characterizes buy the state variables \(P_{i},V_{i}\) and \(T_{i}\) and at some later time in a final in a final state \(f\), the state variables for these two states are related by

\[\frac{P_{f}V_{f}}{T_{f}} = \frac{P_{i}V_{i}}{T_{i}}\ \text{~}\text{(ideal gas in a }\text{scaled}\text{ container)}\text{~}\]

The mathematical relation between the state variables of a system is called the equation of state. Ideal gas will always follow ideal gas equation.

Symbols and constants :

\(P =\) Pressure of gas ; \(\ V =\) Volume of gas ; \(\ T =\) Temperature of gas
\(n =\) no. of moles ; \(\ M =\) mol. wt. of gas; \(\ m_{0} =\) mass of each atom or molecule
\(N =\) total no. of molecule; \(\ N_{0} =\) Avogadro no.
\(K =\) Boltzmann gas constant \(= 1.38 \times 10^{- 23}{JK}^{- 1}\)
R = Universal gas constant
\(\mu =\) Specific gas constant \(= \frac{R}{M};\ m =\) mass of gas

Different forms of Ideal gas of equation

(a) \(\ PV = nRT\)
(b) \(\ P = \frac{dRT}{M}\ \) (d is density of gas)
(c) \(\ PV = NKT\ \) (per molecule)
(d) \(\ PV = m\mu T\)

• \(\ n = \frac{m}{M} = \frac{N}{N_{0}}\)

• \(\ M = m_{0}{\text{ }N}_{0}\)

• \(K = \frac{R}{N_{0}}\)

If mass of the gas is not constant then we can use

\[\frac{P_{1}V_{1}}{n_{1}T_{1}} = \frac{P_{2}V_{2}}{n_{2}T_{2}} = R\text{~}\text{(Universal gas constant)}\text{~}\]

At constant temperature, volume of a fixed mass of a gas is inversely proportional to its pressure.

(a) PV \(=\) constant

(b) \(P = \frac{\text{~}\text{constant}\text{~}}{V}\)

(c) \(V = \frac{\text{~}\text{constant}\text{~}}{P}\)

(d) \(logP =\) constant \(- logV\)

\[\begin{matrix} V \propto \frac{1}{P} \\ PV = \text{~}\text{constant}\text{~} \\ P_{1}{\text{ }V}_{1} = P_{2}{\text{ }V}_{2} \end{matrix}\]

( \(T,n\) are constant)

When a gas is heated at constant pressure, its volume is a linear function of the temperature and can be expressed by the equation for a straight line

\[V = mt + C\]

Where \(t\) is the temperature in \(\ ^{\circ}C\) and \(m\) and \(C\) are constants. The intercept on the vertical axis, \(C\), is \(V_{0}\) which is the volume at \(t = 0^{\circ}C\). The slope of the line is \(m = \frac{\Delta V}{\Delta t}\)
Thus

\[\begin{matrix} & V_{t} = V_{0} + \left( \frac{\Delta V}{\Delta t} \right)t\ (n,p\text{~}\text{are constant}\text{~}) \\ & \frac{\Delta V}{\Delta t} = \text{~}\text{increase in volume per degree}\text{~} \\ & \alpha = \frac{1}{{\text{ }V}_{0}}\frac{\Delta\text{ }V}{\Delta t} = \text{~}\text{relative increase in volume per degree}\text{~} \end{matrix}\]

Thus

\[V_{t} = V_{0} + V_{0}\alpha t = V_{0}(1 + \alpha t)\]

\(\alpha\) is called coefficient of expansion. It is approximately \(\frac{1}{273}\) for all the gases.

\[V = V_{0}\left( 1 + \frac{t}{273} \right)\]

Thus, an increases the temperature of a fixed volume of a gas at constant pressure increases the volume by \(\frac{1}{273}\) of the volume at \(0^{\circ}C\).

\[\frac{V}{V_{0}} = \frac{T}{T_{0}} \]( \(n,P\) are constants)

When the temperature of a gas is changed keeping the volume constant, the pressure of the gas changes. Similar to volume, the pressure changes by \(1/273\).

T

T

\(n,V\) are constants)

( \(n,V\) are constants)
or

\[\begin{matrix} & \beta = \frac{1}{P_{0}}\left\lbrack \frac{\Delta F}{\Delta t} \right\rbrack = \frac{1}{273} \\ & P_{t} = P_{0}\left\lbrack 1 + \frac{t}{273} \right\rbrack = P_{0}\left( \frac{273 + t}{273} \right) \end{matrix}\]

This law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the pressure which each component would exert if placed independently in the container.

\[\begin{matrix} P_{\text{total}\text{~}} = P_{1} + P_{2} + P_{3} + \ldots.. & \ (T,V\text{~}\text{are constants}\text{~}) \\ P_{\text{total}\text{~}} = \sum_{i}^{}\mspace{2mu}\mspace{2mu} P_{i} & \end{matrix}\]

Where the symbol \(\sum_{i}^{}\mspace{2mu}\) stands for the summation over all the components present in the mixture.
The partial pressure \(P_{i}\) of component \(i\) is defined as the pressure that the gas would exert if it were present along in the same volume and the same temperature.

\[\begin{matrix} & p_{1} = \frac{n_{1}RT}{\text{ }V} \\ & p_{2} = \frac{n_{2}RT}{\text{ }V} \end{matrix}\]

The total pressure of the system can be written as

\[\begin{matrix} & P_{\text{total}\text{~}} = \sum_{i}^{}\mspace{2mu}\mspace{2mu} p_{i} \\ & \ = \frac{RT}{V}\sum_{i}^{}\mspace{2mu}\mspace{2mu} n_{i} = n_{\text{total}\text{~}}\frac{RT}{V} \end{matrix}\]

and partial pressure of \(i^{\text{th}\text{~}}\) component can be written as

\[P_{i} = P_{\text{total}\text{~}}\frac{n_{i}}{n_{\text{total}\text{~}}}\]

When \(\frac{n_{i}}{n_{\text{total}\text{~}}}\) is the mole fraction of the respective component

The equation of state for an ideal gas :

\[PV = nRT\]

In this expression, known as the ideal gas law, R is a universal constant that is the same for all gases and T is the absolute temperature in kelvins. Experiments on numerous gases show that as the pressure approaches zero, the quantity \(PV/nT\) approaches the same value R for all gases. For the reason, R is called the universal gas constant. In SI units, in which pressure is expressed in pascals ( \(1\text{ }Pa = 1\text{ }N/m^{2}\) ) and volume in cubic meters, the product PV has units of newton meters, or joules, and \(R\) has the value

\[R = 8.315\text{ }J/mol\text{ }K\]

If the pressure is expressed in atmospheres and the volume in liters \(\left( 1\text{ }L = 10^{3} = 10^{- 3}{\text{ }m}^{3} \right)\), then R has the value

\[R = 0.08214\text{ }L\text{ }atm/mol\text{ }K\]

Using this value of R we find that the volume occupied by 1 mol of any gas at atmospheric pressure and at \(0^{\circ}C(273\text{ }K)\) is 22.4 L .

1. All gases are made of molecule moving randomly in all directions.

2. The size of a molecule is much smaller than the average separation between the molecules.

3. The molecules exert no force on each other or on the walls of the container except during collision.

4. All collisions between two molecules or between a molecule and a wall are perfectly elastic. Also, the time spent during a collision is negligibly small.

5. The molecules obey Newton's laws of motion.

6. When a gas is left for sufficient time, it comes to a steady state. The density and the distribution of position, direction and time. This assumption may be justified if the number of molecules is very large.

Calculation of the pressure of an ideal gas

Consider an ideal gas enclosed in a cubical vessel of edge L . Take a corner of the vessel as the origin O and the \(X - ,Y - ,Z\)-axes along the edges. Let \(A_{1}\) and \(A_{2}\) be the parallel faces perpendicular to the X -axis. Consider a molecule moving with velocity \(\overrightarrow{v}\). The components of the velocity along the axes are \(v_{x},v_{y}\) and \(v_{z}\). When the molecule collides with the force \(A_{1}\), the \(x\)-component of the velocity is reversed whereas the \(y\) - and the \(z\)-components remain unchanged. This follows from our assumption that the collisions of the molecules with the wall are perfectly elastic. The change in momentum of the molecule is

\[\Delta p = \left( - mv_{x} \right) - \left( mv_{x} \right) = - 2mv_{x}.\]

As the momentum remains conserved in a collision, the change in momentum of the wall is

\[\begin{array}{r} \Delta p^{'} = 2{mv}_{x}\#(i) \end{array}\]

After rebound, this molecule travels towards \(A_{2}\) with the x -component of velocity equal to \(- v_{x}\). Any collision of the molecule with any other face (except for \(A_{2}\) ) does not change the value of \(v_{x}\). So, it travels between \(A_{1}\) and \(A_{2}\) with a constant x -component of velocity which is equal to \(v_{x}\). Note that we can neglect any collision with the other molecules in view of the last assumption discussed in the previous section.

The distance travelled parallel to the \(x\)-direction between \(A_{1}\) and \(A_{2} = L\). Thus, the time taken by the molecule to go from \(A_{1}\) to \(A_{2} = L/v_{x}\). The molecule rebounds from \(A_{2}\), travels towards \(a_{1}\) and collides with it after another time interval \(L/v_{x}\). Thus, the time between two consecutive collisions of this molecule with \(A_{1}\) is \(\Delta t = 2L/v_{x}\). The number of collisions of this molecule with \(A_{1}\) in unit time is

\[\begin{array}{r} n = \frac{1}{\Delta t} = \frac{v_{x}}{2\text{ }L}\#(ii) \end{array}\]

The momentum imparted per unit time to the wall by this molecule is, from (i) and (ii), The momentum imparted per unit time to the wall by this molecule is from (i) and (ii),

\[\begin{matrix} & \Delta F = n\Delta p^{'} \\ & \ = \frac{v_{x}}{2\text{ }L} \times 2{mv}_{x} = \frac{m}{\text{ }L}v_{x}^{2} \end{matrix}\]

This is also the force exerted on the wall \(A_{1}\) due to this molecule. The total force on the wall \(A_{1}\) due to all the molecules is

\[\begin{matrix} & & F = \sum_{}^{}\ \frac{m}{\text{ }L}v_{x}^{2} \\ & \ = \frac{m}{\text{ }L}\Sigma v_{x}^{2} & \text{(iii)} \end{matrix}\]

As all directions are equivalent, we have

\[\Sigma v_{x}^{2} = \Sigma v_{y}^{2} = \Sigma v_{z}^{2}\]

\[\begin{matrix} & \ = \frac{1}{3}\Sigma\left( v_{x}^{2} + v_{y}^{2} + v_{z}^{2} \right) \\ & \ = \frac{1}{3}\Sigma v^{2} \end{matrix}\]

Thus, from (iii), \(F = \frac{1}{3}\frac{\text{ }m}{\text{ }L}\Sigma v^{2}\).
If N is the total number of molecules in the sample, we can write

\[F = \frac{1}{3}\frac{mN}{\text{ }L}\frac{\Sigma v^{2}}{\text{ }N}\]

The pressure is force per unit area so that

\[\begin{matrix} & p = \frac{F}{{\text{ }L}^{2}} \\ & \ = \frac{1}{3}\frac{mN}{{\text{ }L}^{3}}\frac{\Sigma v^{2}}{\text{ }N} \\ & \ = \frac{1}{3}\frac{M}{{\text{ }L}^{3}}\frac{\Sigma v^{2}}{\text{ }N} = \frac{1}{3}\rho\frac{\Sigma v^{2}}{\text{ }N} \end{matrix}\]

where \(M\) is the total mass of the gas taken and \(\rho\) is its density. Also \(\Sigma v^{2}/N\) is the average of the speeds squared. It is written as \(u^{2}\) and is called mean square speed. Thus, the pressure is

\[\begin{array}{r} p = \frac{1}{3}\rho v^{2}\#(1) \end{array}\]

or \(\ pV = \frac{1}{3}Mv^{2}\)
or, \(\ pV = \frac{1}{3}Nmv^{2}\)

RMS Speed

The square root of mean square speed is called root-square speed or rms speed. It is denoted by the symbol \(v_{ms}\) Thus,

\[v_{ms} = \sqrt{\Sigma v^{2}/N}\]

or,

\[v^{2} = \left( v_{rm} \right)^{2}\]

Equation (1) may be written as

\[p = \frac{1}{3}\rho v_{rms}^{2}\]

so that

\[v_{rms} = \sqrt{\frac{3p}{\rho}} = \sqrt{\frac{3pV}{M}} = \sqrt{\frac{3RT}{M}}\ (M \rightarrow \text{~}\text{molecular weight}\text{~})\]

Avg. speed \(\ v_{\text{avg}\text{~}} = \sqrt{\frac{8KT}{\pi m_{0}}} = \sqrt{\frac{8RT}{\pi M}}\)
Most probable speed \(v_{p} = \sqrt{\frac{2KT}{m_{0}}} = \sqrt{\frac{2KT}{M}}\)

Translational kinetic energy of gas :

Kinetic energy of any molecule is \(k_{1}\)

\[k_{1} = \frac{1}{2}{\text{ }m}_{0}{\text{ }V}_{1}^{2}\]

Total K.E. of all molecular is k

\[\begin{matrix} k = \Sigma\frac{1}{2}{\text{ }m}_{0}{\text{ }V}^{2} & \\ = \frac{1}{2}{\text{ }m}_{0}\Sigma{\text{ }V}^{2} & = \frac{1}{2}{\text{ }m}_{0}\text{ }N\left( \frac{\Sigma{\text{ }V}^{2}}{\text{ }N} \right) \\ = \frac{m}{2}{\text{ }V}_{rms}^{2} & = \frac{m}{2} \times \frac{3RT}{M} \\ k = \frac{3nRT}{2} & \left( \frac{\text{ }m}{M} = n \right) \\ k = \frac{3PV}{2} & \\ & \text{~}\text{K.E.}\text{~}/\text{~}\text{volume}\text{~} = \frac{3}{2}P \end{matrix}\]

Degree of freedom :

No of ways in which molecule can passes energy is known as degree of freedom.

Law of equipartition of energy :

Statement : For an ideal gas average energy associated with its any molecule for each degree of freedom is \(\frac{KT}{2}\). (Where temperature T in kelvin)
Let \(f\) be the degree of freedom for a gas. Average energy associated with its any molecule \(= \frac{f\text{~}\text{KT}\text{~}}{2}\)
Total kinetic energy of a gas \(= \frac{NfKT}{2}\)
Kinetic energy of one mole of gas \(= \frac{fRT}{2}\)

Value of internal energy of a gas

Internal energy of gas should be sum of K.E. and P.E. of its constitute molecules.
But for ideal gas we has assumed P.E. \(= 0\) (Since force of interaction between molecules is zero)
Thus, I.E. of a gas \(=\) K.E. of molecules
If \(f\) is the degree of freedom of a gas molecules than total K.E. [Trans + Rotation of each molecule] for one mole \(\ U = \frac{fRT}{2}\)
for n mole of gas and having f degree of freedom are at temperature \(T_{1}\) kelvin and heated to temperature \(T_{2}\) then
at \(T_{1}\) kelvin \(\ U_{1} = \frac{nfRT_{1}}{2}\)
at \(T_{2}\) kelvin \(U_{2} = \frac{nfRT2}{2}\)
change in I.E. \(\ \Delta U = \frac{nfR}{2}\left( T_{2} - T_{1} \right)\)

Change in internal energy of ideal gas is a function of temperature and temperature only. It does not depend on how we carry out this change in temperature.
\(\frac{fR}{2}\) is represented as \(C_{v}\)

\[\begin{matrix} & C_{v} = \frac{fR}{2} \\ & U = nC_{v}T \end{matrix}\]

(i) For monoatomic gas \((f = 3)\)

\[U = \frac{n3RT}{2}\ \&\Delta U = \frac{n3R\Delta\text{ }T}{2}\]

or

\[C_{v} = \frac{3R}{2}\]

(ii) For rigid diatomic molecule \((f = 5)\)

\[\begin{matrix} & U = \frac{n5RT}{2}\ \&\ \Delta U = \frac{n5R\Delta\text{ }T}{2} \\ & C_{v} = \frac{5R}{2} \end{matrix}\]

Calculation of work done by gas :

If the piston moves towards right a dx distance then work done by this force is dW.

\[\begin{matrix} & dW = F \cdot dx \\ & \ = P(\text{ }A \cdot dx) \\ & dW = PdV \\ & dW = \int_{V_{1}}^{V_{2}}\mspace{2mu}\mspace{2mu} PdV \end{matrix}\]

Graphical interpretation :

From above integral it can be understood that area enclosed by PV-curve and V-axis represents the work done by gas.

(i)

(ii)

(iii)

(iv)

When volume decrease \(\Delta W\) is -ve

Indirect technique(for quasi-static process) :

Work done by gas + work done by all external agents \(= 0\)
\(\Rightarrow \ \) Work done by gas \(= -\) (work done by all the external agents)

Work done by gas + WD by spring + W.D. by atm press \(= 0\)
W.D. by gas \(= - (\) W. D. by atm press + W.D. by spring)

\[\begin{matrix} & \ = - \left\lbrack \left( - P_{0}Ax \right) + \left( - 1/2{kx}^{2} - 0 \right) \right.\ \\ & \ = P_{0}Ax + \frac{1}{2}{kx}^{2} \end{matrix}\]