When a gas is heated, both the volume and pressure change appreciably. Heat may be supplied to the gas keeping either its volume constant or its pressure constant. The amount of heat required will be different in these two cases for the same rise of temperature. Hence a gas has two specific heat capacities (specific heats).
The specific heat of a gas at constant volume (cv)
It is the amount of heat required to raise the temperature of unit mass of the gas through one kelvin keeping its volume constant.
Molar specific heat at constant volume (Cv)
It is the amount of heat required to raise the temperature of 1 mole of the gas through one kelvin at constant volume.
The specific heat of a gas at constant pressure (cp)
It is the amount of hem required to raise the temperature of unit mass of the gas through one kelvin keeping pressure constant.
Molar specific heat at constant pressure (Cp)
It is the amount of heat required to raise the temperature of 1 mole of the gas through one kelvin at constant pressure.
Unit of specific heat: Jkg- 1 K-1 or J-1 mol-1 K-1
Why Cp is greater than Cv?
When unit mass, one mole, of a gas is heated at constant volume, the heat supplied is utilized only to increase the internal energy of the gas. When it is heated at constant pressure, the heat supplied is used not only for increasing the internal energy but also for doing external work during expansion. For the same rise of temperature, the increase in internal energy is the same in both cases. Hence Cp is greater than Cv.
To show that Cp - Cv = R (Mayer's relation)
Consider one mole of an ideal gas enclosed inside a non-conducting cylinder fitted with a light smooth piston. Let P he its pressure, V its volume and T its temperature.
Keeping the volume constant let the gas be heated so as to raise its temperature by one kelvin. The quantity of heat supplied,
Q = Cv
The amount of heat is used only to raise, the internal energy of the gas.
Now, let us imagine the gas be heated at constant pressure so as to raise the temperature by 1 K. The quantity of heat supplied,
Q’ = Cp
This amount of heat is partly used to raise the internal energy of the gas as the temperature rises by 1 K and also used to do external work as the gas expands to keep the pressure constant.
Therefore, Cp = Cv+ External work
When the gas expands keeping pressure constant, let the piston be moved through a distance dx. Hence external work done by the gas,
W = F x S = PA x dx =PdV;
where A is the area of cross-section of the piston and A x dx = dV, the increase in volume of the gas. Hence equation (iii) becomes,
Cp = Cv + PdV
For an ideal gas,
PV = RT (a)
After expansion, P(V + dV) = R(T + 1) (b)
Eqn: (b) - Eqn: (a) PdV = R
Substituting in Eqn,
Cp = Cv + R;
Therefore, Cp - Cv =R
This is Mayer's equation.
Ratio of specific heats ( γ)
The ratio of specific of a gas is the ratio of its specific heat at constant pressure to the specific heat at constant volume of the gas.
γ =Cp1C,
It is a constant for a gas. γ = 1.4 for diatomic molecule.
Experimental results
Type of gas | Gas | Cp/R | Cv/R | (Cp - Cv)/R | |
1 | Mono-atomic | He A | 2.5 2.5 | 1.5 1.5 | 1.00 1.00 |
2 | Diatomic | H2 O2 | 3.45 3.53 | 2.45 2.53 | 1.00 1.00 |
3 | Polyatomic | CO2 SO2 | 4.43 4.85 | 3.42 3.11 | 1.01 1.08 |
The experimental results show that the result Cp - Cv = R is true for all gases. But the values CP = 2.5R and Cv = 1.5 R is true only for monoatomic gases. Cp and Cv are considerably higher than theoretical values for diatomic and polyatomic gases. This means that these molecules have internal motions with associated energy.
Degrees of freedom
The degrees of freedom of a dynamical system are defined as the total number of independent quantities required to describe the position or motion of the system completely. To understand the concept of degrees of freedom consider the following example.
1. If we consider an ant moving along the straight edge of a table, we can specify its position by knowing the displacement along the line from one end of the table. We say that the ant has only one degree of freedom.
2. If consider the ant moving over a table, to locate its position completely at any instant we must specify the displacements along X and Y axes. Thus, in this case, we say the ant has two degrees of freedom.
3. If we consider a bee flying in space, its position at any instant can be determined by knowing the displacements along the three principal directions i.e., X, Y and Z. Hence the bee has three degrees of freedom.
Degrees of freedom of a gas molecule
1. Monoatomic molecule
The molecule of a mono-atomic gas such as helium, neon, argon etc has only one atom. It is capable of only translatory motion in free space. Hence it has three degrees of freedom.
2. Diatomic molecule
The molecule of a diatomic gas like H2, O2 etc are made up of two atoms joined rigidly through a bond. The molecule can have three translational motions as usual; but in addition, it can rotate about any one of the co-ordinate axes. However; its moment of inertia about the axis joining the two atoms is negligible. So it can have practically two rotational motions. Hence, a diatomic molecule has five degrees of freedom.
3. Triatomic molecule
A molecule such as CO2, NH2, SO2 etc can rotate about any of the three co-ordinate axes. Hence it has six degrees of freedom; three translational and three rotational.
Law of equipartition of energy
For a dynamical system in thermal equilibrium, the total energy of the system is equally divided among the various degrees of freedom.
The share of energy for each degree of freedom = 1/2 kT; where k is the Boltzmann constant and T the temperature.
Specific heats and degrees of freedom
1. Monoatomic gas
A mono-atomic gas molecule has three degrees of freedom
Energy per degree of freedom = (1 /2) k T
Therefore, Energy associated with a molecule = 3 x ½kT
Energy associated with a mole of the gas, E = 3 x (1/2) kT x N = (3/2)RT
Cv= (dE/dT)= (3/2) R; Cp = Cv + R = (3/2)R + R = (5/2)R
γ = Cp/Cv = (5/2)R/(3/2)R = 5/3 = 1.67
2. Diatomic gas
A diatomic molecule has five degrees of freedom.
Energy associated with a molecule = 5 x ½ kT
Energy associated with a mole of the gas = 5 x ½ kT x N = (5/2) RT
Cv = (dE/dT)= (5/2)R; Cp = Cv + R = (5/2)R + R = (7/2) R
γ = cp/cv = (7/2) R (1) (5/2) R = 7/5 = 1.4
3. Triatomic gas
A triatomic molecule has 6 degrees of freedom.
Energy associated with a molecule = 6 x ½ kT
Energy associated with a mole of the gas = 6 x ½ kT x N= 3RT
Cv = (dE /dT) = 3R;
Cp = Cv + R = 3R + R = 4R
γ = Cp/Cv = 4R /3R = 4/3 = 1.33
Note: In general if n is the number of degrees of freedom,
γ = 1 + (2/n)