Question

What is the relation between ${C_p}$ and ${C_v}$ ?

Answer 1

The heat capacity of a substance is defined as the amount of heat which is required to change the temperature of substance by one degree. ${C_p}$ is the molar heat capacity when the pressure is kept constant and ${C_v}$ is the molar heat capacity when the volume of the change is kept constant. In this question, we will define the relation between these two heat capacities.

Complete answer:
We can define heat capacity as the amount of heat which is required to change the temperature of a substance by one degree. There are two types of heat capacities which are ${C_p}$ and ${C_v}$ . ${C_p}$ is the molar heat capacity when the pressure is kept constant and ${C_v}$ is the molar heat capacity when the volume of the change is kept constant.
Now, let us consider one mole of an ideal gas is enclosed in a cylinder with a frictionless piston. So, P, V and T are the pressure, volume and absolute temperature of gas.
A quantity of heat which is dQ is supplied to the gas. To keep the volume of the gas constant, a small weight is placed over the piston. The temperature and the pressure of the gas increase to $T + dT\;$ and $P + dP$ respectively.
The change in internal energy becomes:
$dQ = dU = 1{\text{ }} \times {\text{ }}{C_v}\; \times {\text{ }}dT \ldots ...{\text{ }}\left( 1 \right)$
The additional weight is now removed and the piston moves upwards. The temperature of gas decreases due to expansion of the gas.
Now, heat dQ’ is supplied to the gas till its temperature becomes $T + dT\;$ .
The expansion takes place at constant pressure,
$dQ{\text{ }}\prime = {C_p}dT$
Put the value in first equation:
${C_p}dT = {C_v}dT + dW \ldots ...{\text{ }}(2)\;$
$Work{\text{ }}done,dW = force \times dis\tan ce$
$= {\text{ }}P{\text{ }} \times {\text{ }}A{\text{ }} \times {\text{ }}dx$
Now, the second equation becomes:
${C_p}dT = {C_v}dT + P{\text{ }}dV \ldots ...{\text{ }}\left( 3 \right)$
The equation of ideal gas is for one mole of a gas:
$PV = RT$
On differentiating both the sides, we get
$PdV = RdT\; \ldots ...{\text{ }}\left( 4 \right)$
Now, we substitute equation four in third:
${C_p}dT = {C_v}dT + RdT$
Now, the equation becomes:
${C_p}\; = {C_v}\; + R$
Therefore, the relation between ${C_p}$ and ${C_v}$ is:
${C_p}\; = {C_v}\; + R$

Note :
The value of ${C_p}$ (heat capacity at constant pressure) is always greater than the value of ${C_v}$ (heat capacity at constant volume). This happens because at constant pressure heat is absorbed for increasing internal energy as well as for doing work while in case of constant volume the heat is absorbed only to increase the internal energy of the system.