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```Specific Heat Capacity and Its Relation with Energy

Table of Content

Specific Heat Capacity of Gases

Specific Heat Capacity at Constant Volume (cv)

Specific Heat Capacity at Constant Pressure (cp)

Relation of Cv With Energy

Mono-Atomic Gas (3 Degree of Freedom)

Diatomic Gas

Difference Between two Specific Heat Capacities – (Mayer’s Formula)

Solved Examples

Related Resource

Specific Heat Capacity of Gases

Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of a unit mass of substance through 1ºC.

While defining the specific heat capacity, it was assumed that whole of the heat supplied to the substance is used in raising its temperature and in another way. This assumption holds good in the case of solids and liquids since expansion, in them due to 1ºC rise of temperature is negligible. Gases expand quite appreciably due to a rise in temperature. Therefore, heat required by the gas to do external work cannot be neglected.

A gas can be heated in two ways. Accordingly, there are two specific heat capacities in case of a gas.

(a) Specific Heat Capacity at Constant Volume (cv)

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.

Problem 2:-

Calculate the total rotational kinetic energy of all the molecules in one mole of air at 25.0ºC.

Concept:-

Let us assume all the molecules in the air are diatomic. For half rotation the molecule will contribute ½ kT to its rotational kinetic energy, where k is the Boltzmann’s constant. So for one complete rotation the molecule will contribute kT (½ kT×2= kT) to its rotational kinetic energy.

So the total rotational kinetic energy will be,

E = kT

But k= R/NA

Where R is the gas constant and NA is the Avogadro’s number.

Solution:-

To find out total rotational kinetic energy E, substitute R/NA for k in the equation E = kT,

E = kT

= (R/NA) T

For one mole of gas (NA =1), the equation E =(R/NA)T  will be,

E = RT

To find out the total rotational kinetic energy of all the molecules, substitute 25.0° C for temperature T and 8.31 J/ K for R for one mole of gas in the equation E = RT,

E = RT

= (8.31 J/ K) (25.0° C)

= (8.31 J /K)(25+273) K

= (8.31 J/ K)(298 K)

= 2476.38 J

Rounding off to three significant figure, the total rotational kinetic of all the molecule will be 2480 J.

Problem 3:-

Calculate the internal energy of 1 mole of an ideal gas at 250ºC.

Concept:-

Internal energy (Eint) of the ideal gas is defined as the,

Eint = 3/2 nRT

Where n is the number of moles, R is the gas constant and T is the temperature.

Solution:-

To find out the internal energy of the ideal gas, substitute 1 mole for n, 250° C for temperature T and 8.31 J/ mol. K for the gas constant R in the equation E = 3/2 nRT,

E = 3/2 nRT

= 3/2 (1 mol) (8.31 J/ mol. K) (250° C)

= (1.5)( 1 mol) (8.31 J/ mol. K) (250+273) K

= (1.5)( 1 mol) (8.31 J/ mol. K) (523) K

= 6519.195 J

Rounding off to three significant figures, the internal energy of the ideal gas will be 6520 J.

Problem 4:-

A cosmic-ray particle with energy 1.34 TeV is stopped in a detecting tube that contains 0.120 mol of neon gas. Once this energy is distributed among all the atoms, by how much is the temperature of the neon increased?

Concept:-

We can assume neon is an ideal gas. The internal energy (ΔEint) of the ideal gas is defined as the,

ΔEint = 3/2 nR ΔT

Where n is the number of moles, R is the gas constant and ΔT is the increase in temperature.

From the equation ΔEint = 3/2 nR ΔT, the increase in temperature ΔT will be,

ΔT = 2/3 ΔEint/nR

Solution:-

To obtain the increase in temperature ΔT, substitute 1.34 TeV for ΔEint, 0.120 mol for number of mole n and 8.31 J/ mol. K for the gas constant R in the equation ΔT = 2/3 ΔEint/nR,

ΔT = 2/3 ΔEint/nR

= 2/3 (1.34 TeV) (1012 eV/1 TeV) (1.6×10-19 J/eV) / (0.120 mol) (8.31 J/ mol. K)

= 1.43×10-7 K

From the above observation we conclude that, the increased temperature of the neon will be 1.43×10-7 K.

Problem 5:-

The mass of a helium atom is 6.66×10-27 kg. Compute the specific heat at constant volume for helium gas (in J/kg.K) from the molar heat capacity at constant volume.

Concept:-

Molar heat capacity CV at constant volume of a monoatomic gas for 1 mole is,

CV = 3/2 R

Here R (R = 8.31 J/mol. K) is the gas constant.

Molecular mass of helium is 4.00 g.

Solution:-

We have to find out the specific heat at constant volume for helium gas from the molar heat capacity at constant volume.

Since the molecular mass of helium is 4.00 g, therefore the specific heat at constant volume for helium gas would be,

(CV)H = (3/2 R) / (4.00 g/mol)

= (3/2 ×8.31 J/mol. K) / (4.00 g/mol)

= (3/2 ×8.31 J/mol. K) / ((4.00 g/mol) (10-3 kg/1 g))

= 3116 J/kg. K

Rounding off to three significant figure, the specific heat at constant volume for helium gas would be 3120 J/kg. K.

Problem 5:-

Twelve grams of nitrogen (N2) in a steel tank are heated from 25.0 to 125ºC. (a) How many moles of nitrogen are present? (b) How much heat is transferred to the nitrogen?

Concept:-

Number of moles (n) is equal to given mass of molecule (m) divided by molecular weight (W) of the molecule,

n = m/M

The heat transferred (Q) in a constant volume process is equal to,

Q = nCvΔT

Here Cv is the molar heat capacity at constant volume, ΔT (ΔT =Tf –Ti) is the raising temperature and n is the number of moles.

Thus, Q = nCvΔT

= nCv(Tf –Ti)

Solution:-

To find out the number of moles n of nitrogen, substitute 12 g for given mass m and 28 g/mol for M (molecular weight of nitrogen) in the equation n = m/M,

n = m/M

= (12 g)/(28 g/mol)

= 0.429 mol

Thus 0.429 moles of nitrogen are present.

Since nitrogen (N2) is a diatomic gas, therefore Cv = 5/2 R.

To obtain the heat Q which is transferred to the nitrogen, substitute 0.429 mol for n, 5/2 R for Cv and 125° C for Tf and 25° C for Ti in the equation Q = nCv(Tf –Ti),

Q = nCv(Tf –Ti)

= (0.429 mol) (5/2 R) (125° C-25° C)

= (0.429 mol) (5/2 ×8.31 J/mol. K) ((125 + 273) K-(25 + 273) K)      (Since, R = 8.31 J/mol. K)

= (0.429 mol) (5/2 ×8.31 J/mol. K) (398 K-298 K)

= (0.429 mol) (5/2 ×8.31 J/mol. K) (100 K)

= 891 J

From the above observation we conclude that, the heat Q which is transferred to the nitrogen would be 891 J.

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