Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Specific Heat Capacity and Its Relation with Energy:- 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 (c_{v}):- 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 (C_{v}), is defined as the amount of heat required to raise the temperature of 1 mole of gas through 1ºC keeping its volume constant. C_{v}= Mc_{v} (b)Specific heat capacity at constant pressure (c_{p}):- 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 (C_{p}), 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. C_{p} = Mc_{p} It is self evident from the above discussion that ‘C_{p}’ is greater than ‘C_{v}’ by an amount of heat which is utilized in doing external work. Relation of C_{v} 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] = C_{v} i.e., the heat required to raise the temperature of one mole of gas by 1ºC at constant volume. C_{v} = 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. C_{v }=1/m (dU/dT) = (1/m) (m3N) (1/2 k) = (3/2) R and C_{p} =C_{v}+R = (5/2) R So, γ = C_{p}/ C_{v} = 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 C_{v }= (3/2) R and C_{p} = (5/2) R So, γ = C_{p}/ C_{v} = 5/3 = 1.67 (ii) At medium temperature, the number of degrees of freedom (DOF) is 5. U = (5/2) mRT C_{v }= (5/2) R and C_{p} = (7/2) R So, γ = C_{p}/ C_{v} = 7/5 = 1.4 (iii) At high temperature, the number of degrees of freedom (DOF) is 7. U = (7/2) mRT C_{v} = (7/2) R and C_{p} = (9/2) R So, γ = C_{p}/ C_{v} = 9/7 = 1.29 Difference between two specific heat capacities – (Mayer’s formula):- (a) C_{p }- C_{v} = R/J (b) For 1 g of gas, c_{p }- c_{v} = r/J (c)Adiabatic gas constant, γ = C_{p}/ C_{v} = c_{p}/ c_{v} Solved Examples:- Problem 1:- In an experiment, 1.35 mol of oxygen (O_{2}) 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 = nC_{p}ΔT Here C_{p} is the molar heat capacity at constant pressure, ΔT (ΔT =T_{f} –T_{i}) is the raising temperature and n is the number of moles. Thus, Q = nC_{p}ΔT = nC_{p}(T_{f} –T_{i}) Solution:- Since oxygen (O_{2}) is a diatomic gas, therefore C_{v} = 5/2 R. So, C_{p} = C_{v} + R = 5/2 R +R = 7/2 R At constant pressure, if we double the volume, the temperature will be doubled. T_{i} = 11.0^{°} C = (11.0+273) K = 284 K So, T_{f} = 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 C_{p} , 568 K for T_{f} and 284 K for T_{i} in the equation Q = nC_{p}(T_{f} –T_{i}), Q = nC_{p}(T_{f} –T_{i}) = (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) = 1.12×10^{4} J From the above observation we conclude that, the heat which added to the gas to double its volume would be 1.12×10^{4} 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/N_{A} Where R is the gas constant and N_{A} is the Avogadro’s number. Solution:- To find out total rotational kinetic energy E, substitute R/N_{A} for k in the equation E = kT, E = kT = (R/N_{A}) T For one mole of gas (N_{A} =1), the equation E =(R/N_{A})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 (E_{int}) of the ideal gas is defined as the, E_{int} = 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 (ΔE_{int}) of the ideal gas is defined as the, ΔE_{int} = 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 ΔE_{int} = 3/2 nR ΔT, the increase in temperature ΔT will be, ΔT = 2/3 ΔE_{int}/nR Solution:- To obtain the increase in temperature ΔT, substitute 1.34 TeV for ΔE_{int}, 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 ΔE_{int}/nR, ΔT = 2/3 ΔE_{int}/nR = 2/3 (1.34 TeV) (10^{12} 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 C_{V} at constant volume of a monoatomic gas for 1 mole is, C_{V} = 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, (C_{V})_{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 (N_{2}) 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 = nC_{v}ΔT Here C_{v} is the molar heat capacity at constant volume, ΔT (ΔT =T_{f} –T_{i}) is the raising temperature and n is the number of moles. Thus, Q = nC_{v}ΔT = nC_{v}(T_{f} –T_{i}) 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 (N_{2}) is a diatomic gas, therefore C_{v} = 5/2 R. To obtain the heat Q which is transferred to the nitrogen, substitute 0.429 mol for n, 5/2 R for C_{v} and 125° C for T_{f} and 25^{°} C for T_{i} in the equation Q = nC_{v}(T_{f} –T_{i}), Q = nC_{v}(T_{f} –T_{i}) = (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. Related Resources:- Click here for the Detailed Syllabus of IIT JEE Physics. Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam. To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
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.
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 (C_{v}), is defined as the amount of heat required to raise the temperature of 1 mole of gas through 1ºC keeping its volume constant.
C_{v}= Mc_{v}
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 (C_{p}), 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.
C_{p} = Mc_{p}
It is self evident from the above discussion that ‘C_{p}’ is greater than ‘C_{v}’ by an amount of heat which is utilized in doing external work.
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] = C_{v} i.e., the heat required to raise the temperature of one mole of gas by 1ºC at constant volume.
C_{v} = 1/m (dU/dT)
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.
C_{v }=1/m (dU/dT) = (1/m) (m3N) (1/2 k) = (3/2) R
and
C_{p} =C_{v}+R = (5/2) R
So, γ = C_{p}/ C_{v} = 5/3 = 1.67
(i) At very low temperature, the number of degrees of freedom (DOF) is 3.
U = (3/2) mRT
C_{v }= (3/2) R and C_{p} = (5/2) R
(ii) At medium temperature, the number of degrees of freedom (DOF) is 5.
U = (5/2) mRT
C_{v }= (5/2) R and C_{p} = (7/2) R
So, γ = C_{p}/ C_{v} = 7/5 = 1.4
(iii) At high temperature, the number of degrees of freedom (DOF) is 7.
U = (7/2) mRT
C_{v} = (7/2) R and C_{p} = (9/2) R
So, γ = C_{p}/ C_{v} = 9/7 = 1.29
(a) C_{p }- C_{v} = R/J
(b) For 1 g of gas, c_{p }- c_{v} = r/J
(c)Adiabatic gas constant, γ = C_{p}/ C_{v} = c_{p}/ c_{v}
In an experiment, 1.35 mol of oxygen (O_{2}) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?
The heat transferred (Q) in a constant pressure process is equal to,
Q = nC_{p}ΔT
Here C_{p} is the molar heat capacity at constant pressure, ΔT (ΔT =T_{f} –T_{i}) is the raising temperature and n is the number of moles.
Thus, Q = nC_{p}ΔT
= nC_{p}(T_{f} –T_{i})
Since oxygen (O_{2}) is a diatomic gas, therefore C_{v} = 5/2 R.
So, C_{p} = C_{v} + R
= 5/2 R +R
= 7/2 R
At constant pressure, if we double the volume, the temperature will be doubled.
T_{i} = 11.0^{°} C
= (11.0+273) K
= 284 K
So, T_{f} = 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 C_{p} , 568 K for T_{f} and 284 K for T_{i} in the equation Q = nC_{p}(T_{f} –T_{i}),
Q = nC_{p}(T_{f} –T_{i})
= (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)
= 1.12×10^{4} J
From the above observation we conclude that, the heat which added to the gas to double its volume would be 1.12×10^{4} J.
Calculate the total rotational kinetic energy of all the molecules in one mole of air at 25.0ºC.
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/N_{A}
Where R is the gas constant and N_{A} is the Avogadro’s number.
To find out total rotational kinetic energy E, substitute R/N_{A} for k in the equation E = kT,
= (R/N_{A}) T
For one mole of gas (N_{A} =1), the equation E =(R/N_{A})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,
= (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.
Calculate the internal energy of 1 mole of an ideal gas at 250ºC.
Internal energy (E_{int}) of the ideal gas is defined as the,
E_{int} = 3/2 nRT
Where n is the number of moles, R is the gas constant and T is the temperature.
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.
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?
We can assume neon is an ideal gas. The internal energy (ΔE_{int}) of the ideal gas is defined as the,
ΔE_{int} = 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 ΔE_{int} = 3/2 nR ΔT, the increase in temperature ΔT will be,
ΔT = 2/3 ΔE_{int}/nR
To obtain the increase in temperature ΔT, substitute 1.34 TeV for ΔE_{int}, 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 ΔE_{int}/nR,
= 2/3 (1.34 TeV) (10^{12} 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.
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.
Molar heat capacity C_{V} at constant volume of a monoatomic gas for 1 mole is,
C_{V} = 3/2 R
Here R (R = 8.31 J/mol. K) is the gas constant.
Molecular mass of helium is 4.00 g.
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,
(C_{V})_{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.
Twelve grams of nitrogen (N_{2}) 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?
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 = nC_{v}ΔT
Here C_{v} is the molar heat capacity at constant volume, ΔT (ΔT =T_{f} –T_{i}) is the raising temperature and n is the number of moles.
Thus, Q = nC_{v}ΔT
= nC_{v}(T_{f} –T_{i})
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,
= (12 g)/(28 g/mol)
= 0.429 mol
Thus 0.429 moles of nitrogen are present.
Since nitrogen (N_{2}) is a diatomic gas, therefore C_{v} = 5/2 R.
To obtain the heat Q which is transferred to the nitrogen, substitute 0.429 mol for n, 5/2 R for C_{v} and 125° C for T_{f} and 25^{°} C for T_{i} in the equation Q = nC_{v}(T_{f} –T_{i}),
Q = nC_{v}(T_{f} –T_{i})
= (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.
Related Resources:-
Click here for the Detailed Syllabus of IIT JEE Physics.
Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam.
To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
THIRD LAW OF THERMODYNAMICS:- In all heat engines,...
Solved Examples on Thermodynamics:- Problem 1 :- A...
Enthalpy of Reaction It is the enthalpy change...
Enthalpy of a System The quantity U + PV is known...
Specific Heat Capacity or Specific Heat [c]:- It...
Thermodynamic change or Thermodynamic Process:-...
HESS’S LAW This law states that the amount...
Level 2 Objective Problems of Thermodynamics Level...
Work Done During Isothermal Expansion:- Consider...
Introduction to Thermodynamics:- Thermodynamics:-...
Application of Hess's Law 1. Calculation of...
Relationship between free Energy and Equilibrium...
The First Law of Thermodynamics:- The first law of...
Objective Questions of Thermodynamics and Answers...
Work Done During Adiabatic Expansion:- Consider...
Macroscopic Properties He properties associated...
Solved Problems on Specific Heat, Latent Heat and...
Solved Problems on Thermodynamics:- Problem 1:- A...
Thermodynamic State of a System and Macroscopic...
Second Law of Thermodynamics:- Entropy:- The...
Application of bond energies (i) Determination of...
Miscellaneous Exercises Thermal Physics:- Problem:...
Gibbs Free Energy This is another thermodynamic...
Reversible and Irreversible Process:- Reversible...
BOMB CALORIMETER The bomb calorimeter used for...