Specific Heat Capacity & Mean Free Path | Physics for EmSAT Achieve PDF Download

Specific Heat Capacity

Specific Heat Capacity is the amount of energy required by a single unit of a substance to change its temperature by one unit. When you supply energy to a solid, liquid or gas, its temperature changes. This change of temperature will be different for different substances like water, iron, oxygen gas, etc.
This energy is known as the Specific Heat Capacity of the substance and is denoted by ‘C’. Molar Specific Heat Capacity of a substance is C and is calculated for one mole of the substance. Mathematically we can write:
C = ΔQ/m
Further, when you supply energy to a substance, it may undergo a change in volume and/ or pressure, especially in gaseous substances. Hence, to determine the Specific Heat Capacity of gases, it is important to pre-determine the pressure and volume under which you want to calculate C since it can have infinite values (depending on the values of pressure and volume). The Molar Specific Heat Capacity at constant volume is denoted by C_v and that at constant pressure is denoted by C_p.

Specific Heat Capacity of Gases

According to the first law of thermodynamics ΔQ = ΔU + ΔW {change in heat of a system = change in internal energy + amount of work done}. Change in heat of a system (ΔQ) can also be calculated by multiplying Mass (m), Specific Heat Capacity (C) and change in Temperature (ΔT):
ΔQ = mCΔT             Or,
mCΔT = ΔU + ΔW ————(1)

Monatomic Gases (Monoatomic gases)

In monatomic gas, molecules have three translational degrees of freedom. At temperature ‘T’ the average energy of a monatomic molecule is (3/2)K_BT. Now, let’s look at one mole of such a gas at constant volume and calculate the internal energy (U):
U = (3/2) K_BT x N_A {where NA is Avogadro constant}
The total internal energy will by the internal energy of a single molecule multiplied by the number of molecules in one mole of the gas; which is Avogadro constant N_A. Now, Boltzmann’s constant (K_B) is the Gas constant (R) divided by N_A. Hence: U = (3/2)(R/N_A)T × N_A 
U = (3/2)RT ———————(2) 
In equation (1), since the energy is supplied at constant volume: mC_vΔT = ΔU + ΔW. For one mole of a gas, m = 1. Also, for calculating C_v, ΔT = 1. Since the volume is constant, ΔW = 0. Therefore, 1×C_v×1 = ΔU + 0 
C_v = ΔU = (3/2)RT —————-[refer (2)] 
So, the molar specific heat capacity to change the temperature by 1 unit would be C_v = (3/2)R. For an ideal gas, C_p – C_v = R (Gas Constant). Therefore: C_p = R + C_v = R + (3/2)R = (5/2)R. The ratio of C_p:C_v (γ) is hence 5:3. 

Diatomic Gases

In case of diatomic gases, there are two possibilities:

• Molecule is a Rigid Rotator: In this scenario, the molecule will have five degrees of freedom (3 translational and two rotational). As defined by the Law of Equipartition of Energy, the internal energy (U) can be calculated as:

U = (5/2)K_BT * N_A = (5/2)RT. Following the calculation used for monatomic gases: 
C_v = (5/2)R
C_p = (7/2)R
γ = 7:5

• Molecule is NOT a Rigid Rotator: In this scenario, the molecule will have an additional vibrational degree of freedom. The internal energy can thus be calculated as:

U = [(5/2)K_BT + K_BT] * N_A = [(5/2)(R/N_A)T + (R/N_A)T] * N_A = (7/2)RT 
Following the calculation used for monatomic gases:  C_v = (7/2)R 
C_p = (9/2)R and hence γ = 9:7 

Polyatomic Gases

The degrees of freedom of polyatomic gases are:

• 3 translational
• 3 rotational
• f vibrational

Deploying the Law of Equipartition of Energy for calculation of internal energy, we get:

U = [(3/2)K_BT + (3/2)K_BT + fK_BT] * N_A = [(3/2)(R/N_A)T + (3/2)(R/N_A)T + f(R/N_A)T] * N_A 
U = (3 + f)RT 
The molar specific heat capacities: 
C_v = (3+f)R
C_p = (5+f)R

Specific Heat Capacity of Solids

Using the Law of Equipartition of energy, the specific heat capacity of solids can be determined. Let us consider a mole of solid having N_A atoms. Each atom is oscillating along its mean position. Hence, the average energy in three dimensions of the atom would be:
3 * 2 * (1/2)K_BT = 3K_BT 
For one mole of solid, the energy would be: 
U = 3K_BT * N_A = 3(R/N_A)T * N_A = 3RT —————–(3) 
If the pressure is kept constant, then according to the laws of thermodynamics 
ΔQ = ΔU + PΔV 
In case of solids, the change in volume is ~0 if the energy supplied is not extremely high. Hence, 
ΔQ = ΔU + P * 0 = ΔU 
So, the molar specific heat capacity to change the temperature by 1 unit would be: 
C = 3R ——————-[refer (3)] 

Specific Heat Capacity of Water

For the purpose of calculation of specific heat capacity, water is treated as a solid. A water molecule has three atoms (2 hydrogens and one oxygen). Hence, its internal energy would be:
U = (3 * 3K_BT)*N_A = 9K_BT*N_A = 9(R/N_A)T * N_A = 9RT 
And, following a similar calculation like solids: C = 9R 

Mean Free Path

The Kinetic theory of gases assumes that molecules are continuously colliding with each other and they move with constant speeds and in straight lines between two collisions. Hence, a molecule follows a chain of zigzag paths. Each of these paths is known as a free path (since it lies between two collisions).
The average distance travelled by the molecule between two collisions is known as the Mean Free Path. The number of collisions increases if the gas is denser or the molecules are large in size. 

If the molecules of a gas are spheres having a diameter – ‘d’; one molecule is moving with an average speed – ‘v’ and the number of molecules per unit volume is “n”, then the mean free path (l) can be calculated by using the formula: l = 1/πnd²
The equation is arrived at under the assumption that all other molecules are at rest, which is not actually the case. If we consider all molecules to be moving in all directions with different speeds, then the mean free path formula would be: l = 1/√2 πnd²

Solved Question

Q. One mole of an ideal monoatomic gas is mixed with 1 mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume is (in cal):
(a) 22 cal    
(b) 4 cal      
(c) 8 cal        
(d) 12 cal
Ans: (b)
Solution: As we know C_v = (3/2) R for a monoatomic gas and C_v = (5/2) R for a diatomic case.
Thus for the mixture, average of both is = [(3/2) R + (5/2) R] /2 = 2R = 4 cal.