Relation between Cp and Cv - Meyer's formula with Derivation
Cp represent the molar specific heat at constant pressure and Cv represent the molar specific heat at constant volume. so, the relation between Cp and Cv is the difference between Cp and Cv is always equal to the gas constant. that is given by,
Cp − Cv = R
Now, what is meant by Cp, Cv and R?
Cv (Molar specific heat at constant volume):- It is defined as the amount of heat required to increase the temperature of 1 mole of a gas through 1°C at constant volume. it is denoted by Cp.
Cp (Molar specific heat at constant pressure):- It is defined as the amount of heat required to increase the temperature of 1 mole of gas through 1°C at constant volume. it is denoted by Cp.
R (gas constant):- R is the universal gas constant for one mole of gas. As R is always positive, it follows that Cp > Cv.
After knowing the concept of Cp and Cv. we noticed that both Cp and Cv measures the amount of heat which increase the temperature of 1 mole of gas. but the difference is Cp increases the temperature of gas at constant pressure. on the other hand Cv increases the temperature of the gas at constant volume. But both Cp and Cv have many similarities too. Cp and Cv both are the special significance of specific heats of a gas.
Specific heat of a gas
When a gas is heated, its temperature increases due to this the volume and pressure of gas also change. this is the conditions of all types of gas. but why we discuss specific heat. the answer is the amount of heat required to increase the temperature of 1 gram of gas at 1°C is not fixed. even specific heat of any gas is not fixed. a gas can have any value of specific heat. gas always depending on the conditions under which it is heated. So in simple or one word we can say that a gas does not possess a unique specific heat. This can be clearly by seeing the following examples.
Examples of specific heat of gas 'C'
lets consider the mass of a gas be 'm' enclosed in cylinder with air tight and attached with piston.
- Suppose the gas is suddenly compressed. remember no heat is supplied to the gas. i.e., Δ Q = 0. But a little bit of temperature changes due to the suddenly compression of gas.
∴ C = Δ Q ∕ m× ΔT = 0
so, specific heat of the gas is Zero.
- Now, the gas is heated and allowed to expand in such as way that temperature of gas increases due to heat supplied is equal to the decreases in temperature due to expansion of gas itself. then the total temperature is Zero. i.e., ΔT = 0.
∴ C = Δ Q ∕ m× ΔT = ∞
so, the specific heat of the gas is infinite.
- Again, the gas is heated and allowed to expand in such a way that the decrease in temperature due expansion is less than the increase in temperature due to heat supplied. so the temperature of the gas will rise. i.e., ΔT is positive.
∴ C = Δ Q ∕ m× ΔT = positive
so, specific heat of the gas is positive.
- Finally, the gas is heated and allowed to expand in such a way that the decrease in temperature due expansion is greater than the increase in temperature due to heat supplied. so the temperature of the gas will decrease. i.e., ΔT is Negative.
∴ C = Δ Q ∕ m× ΔT = Negative value
so, the specific heat of the gas is Negative
Thus, by the all above example we conclude that the specific heat of a gas may be Negative, Positive, Infinite and Zero. so the exact value of the specific heat depends on the conditions under which gas is heated.
Derivation of relation between Cp and Cv
By applying the First law of Thermodynamics we obtained the relation between Cp and Cv. please note that relation between Cp and Cv may be written as relation between the two specific heats of gas in some case.
Relation between Cp and Cv : Mayer's Relation
consider one mole of and ideal gas. heat the gas so that the temperature increases by ΔT. Now, according to the law of Thermodynamics, the heat supplied ΔQ is used partly to increase the internal energy and partly in doing the work of expansion. that is given by,
ΔQ = ΔU + P ΔV
If the heat ΔQ is observed at constant volume, then ΔV = 0 and we have,
Here we dropped the volume(V) on every quantity because the internal energy U of an ideal gas depends only on its temperature T.
Now, If the heat ΔQ is observed at constant pressure, then
Again, we have dropped the pressure(P) from the first term U of an ideal gas depends only on T. Clearly,
But for one mole of an ideal gas, PV = RT.
Differentiating both side with respect to T for constant pressure P,
Hence, Cp - Cv = R
This is the required relation between Cp and Cv. It is also known as Mayer's formula.
Examples of relation between Cp and Cv.
Calculate the specific heat at constant volume for a gas. Given specific heat at constant pressure is 6.85 cal.mol¯1 Kㄧ1., R = 8.31 J mol¯1 Kㄧ1 and J = 4.18 J cal¯1.
Solution:-
Here Cp = 6.85
R = 8.31, J = 4.18
As Cp
Cp − Cv = R/J
Cv = Cp - R/J = 6.85 - 8.31/4.18
= 6.85 - 1.988 = 4.862 cal.mol¯1 Kㄧ1.
Relation between Cp and Cv for 2 mole of an ideal gas.
For 2 mole of an ideal gas, the relation between Cp and Cv is Cp − Cv = R. because for 2 mole of an ideal gas, 2 is multiplied on both side. this explains as below.
As we know that,
For one mole of gas the relation between Cp and Cv is c. where R is universal gas constant.
For 2 mole of gas,
2Cp − 2Cv = 2R
2 is multiplied by both side,
2(Cp − Cv) = 2R
Hence, both the 2 is canceled by each other. so the final result of relation between Cp and Cv of 2 moles is Cp − Cv = R.
Please Note that:- The above relation will remains same even if we consider any number of moles of gas because in that case both side will get multiplied by the same number.
If any of you have doubt that this is wrong. because everywhere you see the relation between Cp and Cv is Cp − Cv = 2R. but this is wrong.
You can see for your reference in class 11 physics by S.L Arora. page no. 12.11 in Thermodynamics chapter. there is also use the relation Cp − Cv = R. not Cp − Cv = 2R.
If you have any doubt hen comment below.
Ratio of Cp and Cv or Cp/Cv ratio
The ration of two specific heat Cp and Cv is represented by Gamma 𝛄. that is given by,
Cp ∕ Cv = 𝛄
Please Note that the value of gamma depends upon the atomicity of the gas.