Thermodynamics Concept Page - 7

Definition

Specific heat capacity for an adiabatic process

The specific heat capacity for an adiabatic process is zero.

Definition

Specific heat capacity for an isothermal process

c=±∞$c=\pm \mathrm{\infty }$ (the plus sign corresponds to addition of heat to a system, the minus sign indicates the removal of heat from a system). 

Definition

Rotational degree of freedom

A diatomic molecule can rotate about three axes.Multi-atomic gas molecules may undergo rotational motions associated with the structure of the molecule. A monatomic molecule has zero rotational degrees of freedom.

Definition

Vibrational degrees of freedom

There may be intermolecular vibrational motion between nearby gas particles, and vibrational motion arising from intermolecular forces between atoms that form the molecules. A monoatomic molecule has zero vibrational degrees of freedom. A diatomic molecule has one vibrational degree of freedom.

Definition

Cp/Cv for monoatomic,diatomic and polyatomic molecules

Monoatomic gas
CpCv=53${\displaystyle \frac{C_p}{C_v}}={\displaystyle \frac{5}{3}}$
Diatomic gas
CpCv=75${\displaystyle \frac{C_p}{C_v}}={\displaystyle \frac{7}{5}}$
Polyatomic molecules
CpCv=4+f3+f${\displaystyle \frac{C_p}{C_v}}={\displaystyle \frac{4+f}{3+f}}$
where f$f$ is the number of degrees of freedom in molecular motion

Formula

Specific heat capacity at constant volume

Specific heat capacity at constant volume is the amount of heat energy required to raise the temperature of unit mass of the substance by 1 K at constant volume.
(∂U∂T)v=(∂Q∂T)v=Cv${\left({\displaystyle \frac{\mathrm{\partial }U}{\mathrm{\partial }T}}\right)}_v={\left({\displaystyle \frac{\mathrm{\partial }Q}{\mathrm{\partial }T}}\right)}_v=C_v$

Formula

Specific heat capacity at constant pressure

Specific heat capacity at constant pressure is the amount of heat energy required to raise the temperature of unit mass of the substance by 1 K at constant pressure.
(∂H∂T)p=(∂Q∂T)p=Cp${\left({\displaystyle \frac{\mathrm{\partial }H}{\mathrm{\partial }T}}\right)}_p={\left({\displaystyle \frac{\mathrm{\partial }Q}{\mathrm{\partial }T}}\right)}_p=C_p$

Formula

Relation between Cp and Cv

Cp=Cv+R$C_p=C_v+R$ where R$R$ is the universal gas constant.

Definition

Dulong and Petit's Law

The law states that the specific heat of solids is 3RM${\displaystyle \frac{3R}{M}}$, where R is the gas constant$\text{gas constant}$ and M$M$ is the molecular weight$\text{molecular weight}$ of the solid.

Example

Example of Dulong Petit's Law

Calculate the atomic weight of element having specific heat capacity equal to 620$620$ J⋅K−1⋅kg−1$J\cdot K^{-1}\cdot k{g}^{-1}$
Answer: Specific heat of element (c)=3RM$(c)=3{\displaystyle \frac{R}{M}}$, where M$M$ is in kg⋅mole−1$kg\cdot mol{e}^{-1}$
∴M=3Rc$\therefore M=3{\displaystyle \frac{R}{c}}$
M=3×8.314620$M=3\times {\displaystyle \frac{8.314}{620}}$
M=0.04 kg⋅mole−1=40 g⋅mole−1$M=0.04kg\cdot mol{e}^{-1}=40g\cdot mol{e}^{-1}$

Law

Applicationof Dulong Petit's Law

This law is valid only at a higher temperature which varies for different solid elements.
If you look at the graph, you can see that for the lead, the graph is flattening out after 200K temperature and for silicon, it is flattening after 600K. This tells that specific heat capacity of elements becomes constant after certain high temperature. This also proves that law is valid for different higher temperature for various elements.

Example

Problems involving heat capacity

A cylinder of fixed capacity 67.2$67.2$ litres contains helium gas at S.T.P. The amount of heat required to raise the temperature of the gas by 150C${15}^{0}C$ is found as follows:
(R =$=$8.31 J/mol/K$8.31J/mol/K$)Since, the process is at constant volume,
Q=U$Q=U$ as W=0$W=0$
Thus, Q = nCvδT$n{C}_v\delta T$
At STP, n=$n=$ PVRT${\displaystyle \frac{PV}{RT}}$
Since, He is diatomic, Cv=2.5R$C_v=2.5R$
Q = PVRT×2.5R×15${\displaystyle \frac{PV}{RT}}\times 2.5R\times 15$
Substituting the pressure and temperature values at STP, 
P=1 atm$P=1atm$
V=67.2$V=67.2$ L
T=15 0C$T=15{}^{0}C$
We get,
Q=170×102J

BookMarks

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