show that the specific heat capacity of solid is equal to three times that of gas constant (C=3R)
Answers
Specific heat capacity at constant volume is defined as the amount of heat required to raise the temperature of 1 g of the gas through 1ºC keeping volume of the gas constant.
If we take 1 mole of gas in the barrel, the corresponding specific heat capacity is called Gram molar specific heat capacity at constant volume.
Molar specific heat capacity, at constant volume (Cv), is defined as the amount of heat required to raise the temperature of 1 mole of gas through 1ºC keeping its volume constant.
Cv= Mcv
(b)Specific Heat Capacity at Constant Pressure (cp)
Specific heat capacity, at constant pressure, is defined as the amount of heat required to raise the temperature of 1 g of gas through 1ºC keeping its pressure constant.
In case of 1 mole of the gas:
Gram molecular specific heat capacity of a gas (Cp), at constant pressure, is defined as the amount of heat required to raise the temperature of 1 mole of the gas through 1ºC keeping its pressure constant.
Cp = Mcp
It is self evident from the above discussion that ‘Cp’ is greater than ‘Cv’ by an amount of heat which is utilized in doing external work.
Relation of Cv With Energy
From first law of thermodynamics,
(dQ)v = dU
Or (1/m) [(dQ)v/dT] = (1/m) (dU/dT)
By definition (1/m) [(dQ)v/dT] = Cv i.e., the heat required to raise the temperature of one mole of gas by 1ºC at constant volume.
Cv = 1/m (dU/dT)
(a) Mono-Atomic Gas (3 Degree of Freedom)
Total energy, U = mN 3 [(1/2) KT], Here m is the number of moles of the gas and N is the Avogadro’s number.
Cv =1/m (dU/dT) = (1/m) (m3N) (1/2 k) = (3/2) R
and
Cp =Cv+R = (5/2) R
So, γ = Cp/ Cv = 5/3 = 1.67
(b) Diatomic Gas
(i) At very low temperature, the number of degrees of freedom (DOF) is 3.
U = (3/2) mRT
Cv = (3/2) R and Cp = (5/2) R
So, γ = Cp/ Cv = 5/3 = 1.67
(ii) At medium temperature, the number of degrees of freedom (DOF) is 5.
U = (5/2) mRT
Cv = (5/2) R and Cp = (7/2) R
So, γ = Cp/ Cv = 7/5 = 1.4
(iii) At high temperature, the number of degrees of freedom (DOF) is 7.
U = (7/2) mRT
Cv = (7/2) R and Cp = (9/2) R
So, γ = Cp/ Cv = 9/7 = 1.29
Difference Between two Specific Heat Capacities – (Mayer’s Formula)
(a) Cp - Cv = R/J
(b) For 1 g of gas, cp - cv = r/J
(c)Adiabatic gas constant, γ = Cp/ Cv = cp/ cv
Solved Examples
Problem 1:-
In an experiment, 1.35 mol of oxygen (O2) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?
Concept:-
The heat transferred (Q) in a constant pressure process is equal to,
Q = nCpΔT
Here Cp is the molar heat capacity at constant pressure, ΔT (ΔT =Tf –Ti) is the raising temperature and n is the number of moles.
Thus, Q = nCpΔT
= nCp(Tf –Ti)
Solution:-
Since oxygen (O2) is a diatomic gas, therefore Cv = 5/2 R.
So, Cp = Cv + R
= 5/2 R +R
= 7/2 R
At constant pressure, if we double the volume, the temperature will be doubled.
Ti = 11.0° C
= (11.0+273) K
= 284 K
So, Tf = 2(284 K)
= 568 K
To obtain heat which added to the gas to double its volume, substitute 1.35 mol for n, 7/2 R for Cp , 568 K for Tf and 284 K for Ti in the equation Q = nCp(Tf –Ti),
Q = nCp(Tf –Ti)
= (1.35 mol) (7/2 R) (568 K-284 K)
= (1.35 mol) (7/2 ×8.31 J/mol. K) (568 K-284 K) (Since, R = 8.31 J/mol. K)
= 1.12×104 J
From the above observation we conclude that, the heat which added to the gas to double its volume would be 1.12×104 J.
Answer:
W.K.T,
Cp-Cv=R --------1
Now,
Cv=dU/dT (U-internal energy) -------2
U=f/2*R*T (f- degree of freedom)
[So, for solid atom, f=4 ,i.e., 3 for transition and 1 for rotation]
Therefore, U= 4/2*R*T
= 2*R*T
Substituting U in eqn 2, we get
Cv= d/dt (2RT)
Cv= 2R
Now, substituting Cv in eqn 1, we have
Cp-Cv= R
Cp-2R= R
Cp= R+2R
Cp= 3R
Hope, this helps you :-)