In general specific heat(C)  gives us an idea of the amount of energy(heat) we need to provide to a system in order to bring about a unit rise in the temperature of the system. It's value may vary depending on the process you are providing this energy. Hence we have two values of C namely Cv and Cp .

Cv for a gas is the change in internal energy (U) of a system with respect to change in temperature at a fixed volume of the system i.e. Cv =(∂ U/∂ T)v  whereas Cp for a gas is the change in the enthalpy (H) of the system with respect to change in temperature at a fixed pressure of the system i.e Cp = (∂ H/∂ T)p.

We know that, ΔH = ΔU + PΔV (+ VΔP, ΔP=0 for constant pressure) . So the enthalpy term is  greater than the internal energy term because of the PΔV term i.e in case of a constant pressure process more energy is needed, to be provided to the system as compared to that of a constant volume process to achieve the same temperature rise, as some energy is utilized in the expansion work of the system. And the relation that correlates these two is Cp = Cv + R

But since liquids and solids can practically assumed to be incompressible, Cp and Cv for them have almost same values and hence only a single value of specific heat is used for them.

In general specific heat(C)  gives us an idea of the amount of energy(heat) we need to provide to a system in order to bring about a unit rise in the temperature of the system. It's value may vary depending on the process you are providing this energy. Hence we have two values of C namely Cv and Cp .

Cv for a gas is the change in internal energy (U) of a system with respect to change in temperature at a fixed volume of the system i.e. Cv =(∂ U/∂ T)v  whereas Cp for a gas is the change in the enthalpy (H) of the system with respect to change in temperature at a fixed pressure of the system i.e Cp = (∂ H/∂ T)p.

We know that, ΔH = ΔU + PΔV (+ VΔP, ΔP=0 for constant pressure) . So the enthalpy term is  greater than the internal energy term because of the PΔV term i.e in case of a constant pressure process more energy is needed, to be provided to the system as compared to that of a constant volume process to achieve the same temperature rise, as some energy is utilized in the expansion work of the system. And the relation that correlates these two is Cp = Cv + R

But since liquids and solids can practically assumed to be incompressible, Cp and Cv for them have almost same values and hence only a single value of specific heat is used for them.

In addition to Srujan Singh's answer that I fully acknowledge, here is a small comparison for the  p·ΔV term.

Cheers,

Petra

At higher temperatures, the difference between Cp and Cv in solids is not so small. However, it is not easy to measure Cv of solids (because of the increasing pressure). There is an equation to calculate Cv from known Cp:

Cv = Cp - alpha^2 K V T.

Alpha is the thermal expansion coefficient, K is the isothermal bulk modulus and V and T are the volume and the temperature.

because density of a gas changes with pressure

In gas volume. Changes due to temperature different 

Gas are compressible ( there is an effect of P on V). Liquids are not compressible ( P has almost no effect on V)

It is not entirely correct to say that liquids gave only 1 heat capacity. Actually they have both Cp and Cv, whose values are widely reported. However it should be pointed out that on one hand the pertinent measurements can be particularly inexact and on the other hand – in the engineering equipment the actual heat capacity of liquids could be somewhere among Cp and Cv.

due to compressiblity factor.

There is excellent treatment of CP and CV  in Gugenheim for gases and condensed phases (p.105&121),  which shows that the arguments of  Dr. Singh was over simplification for the  gases and it was completely missing for the condensed phases, which may be given by Dr. Benisek properly.  [alpha2  T Vm / X], where  X  is isothermal compresibilty (i.e., inverse  bulk modulus). If one takes the first few  two terms in the power series representation of volume of any phase such as ;

V=  RT/P  +  B  + { C P  +  D P2 ....+ }   namely   V= RT/P + B(T)  one can show that;

alpha=  [dLn Vm /dT] P = R/P  + dB/dT  similarly  x-1 = -[d Ln(Vm /dP)]T =P {1+BP/RT} 

Then one gets   CP  -  CV  =R [1 +  P/ R  dB/dT ]2   where  Vm molar volume

NEW YEAR GREETING

@ S. Singh,

Please allow me to make the following comments regarding your post:

i) the enthalpy-entropy relation you provided applies for constant pressure processes (the system is in mechanical equilibrium with its surrounding). The more general equation is: ΔH = ΔU + Δ(PV) = ΔH = ΔU + PΔV + VΔP

ii) Cp = Cv + nR, and this equation applies for ideal gases. In general, Cp=Cv + a2TV/KT, where a is the expansion coefficient and KT is the isothermal compressibility. This equation is considered universally true.

iii) Liquids and solids may not necessarily have identical/close Cp and Cv.

For example, for water at 25°C, the equation above gives Cp,m = 75.3 J K-1 mol-1 compared with CV,m = 74.8 J K-1 mol-1. In some cases, the two heat capacities differ by as much as 30 per cent. (Ref. Atkins, 8th ed., pp. 63).

Dear Talal,

I absolutely agree with your points. But as you can see my answer is for a much simplified and  ideal case. (And also ΔP=0 for constant pressure process and ΔV=0 for constant volume process, so that reduces the equation and I have just stated for a constant pressure process). But thanks for your corrections.

Dear Dr. Shahwan, What you have done in section ı) is nothing but the unfinished work of Legendre transformation to show that enthalpy is a new characteristic function where S and P are independent variables: To get the proper result one has to write;

dU= TdS - pdV and then  sum up with your expression to obtain: dH= T dS + VdP. Closed System.

For open multicomponent systems one reads:. dH=TdS + VdP + SUMk ChemPotk Nk.

ii) Expression one is alright but Cv and Cp mostly are given in terms of molar quantities. One can also use weight unıts to define. What is n? Is it number of moles or the number of molecules. What is V? Is it molar volume or specific volume?

What do you mean by saying ' The enthalpy - entropy relation you provided applies for constant pressure processes...?

YES ! The expression given by Dr. Singh is falls ΔH = ΔU + PΔV?? And yours is right.

SORRY. But those expressions (yours and Singh's ) have -as  they appear now- nothing to do with H and S connection YET.  Only Helmholtz and Gibbs Free Energies have direct connections with entropy through Legendre Transformation :(Callen)

H=U+PV;   F= U-TS  and G= U+PV-TS  QED.    Best Regards

NEW YEAR GREETING

Thank you.

the answers gives are more exaustive about the question

That is because they are roughly the same for solid or liquid phase.

Dear Dr. Ogurtani

Thank you. I was referring to enthalpy-internal energy relation, not enthalpy-entropy relation as mistakenly written in the earlier post.

Happy New Year

simply because liquids and solids are considered incompressible

Dear Lioua, how about those hypothetic solids , which are  incompressible but they can go though the shear  deformation under the external tractions and even body forces!  Can you say now  that the  Gibbs free energy is replacable by the Helmholz free enerğy to discuss the behaviour of the system under the isothermal and  constant tractions and body forces?   

By basic definition of specific heat: amount heat added to cause a unit rise in temperature per unit mass. In solids addition of heat results in no appreciable change in volume and quantity of heat remains nearly same for constant  volume or pressure condition.

Where as in gases the quantity of heat varies depending on constant pressure or constant volume conditions. Therefore we have two specific heats for gases i.e. Cp and Cv.

Always Cp > Cv as some of the heat supplied under constant pressure is transformed in to work (P.dv) where as under constant volume condition all the heat added is stored as internal energy which is limited.

dear Tilak,

as dr Benisek and dr Shyu said, CV and CP are (approximately) the same for liquids and solids because matter in these conditions is usually considered incompressible:

CP = CV + T alfa_P^2 V/k_T

T = temperature, alfa_P = coefficient of isobaric expansion, k_T = coefficient of isothermal compressibility, V = specific volume

costante

The Mayer's relation provides clear answer to your question. From a thermodynamic point of view, to better understand it, you can read the following link:

http://en.wikipedia.org/wiki/Relations_between_heat_capacities

You can accept the approximation Cp = Cv if you can accept the apporximation that volume (at constant P) do not change with Temperature, or that pressure, at constant V, do not change with Temperature. 

These arguments are normally found on basic courses of thermodynamics, during the first years of any scientific or thechnical university studies. Anyone with a scientific degree should know the answer, or at least be able to find it by himself..

I agree with you, LUCIO, this can be found in any basic thermodynamic course.

The molar enthalpy is given by:             H ̃= U ̃+ PV ̃

The molar volumes of liquids and solids do not change very much when the temperature or pressure change by small to moderate amounts.

Therefore, it is a reasonable approximation to treat the molar volume as a constant as long as the temperature and pressure don’t change by too much;  we obtain the followings:

                            ∂H ̃= ∂U ̃+ ∂(PV ̃ ) ≈ ∂U ̃+ V ̃∂P

Since the molar volume of solids and liquids is quite very small,

                                              ∂H ̃≈ ∂U ̃

or                                   (∂H ̃)/∂T≈ (∂U ̃)/∂T

The Constant Volume Heat Capacity is defined by:       Cv = ((∂U ̃/∂T)V 

The Constant Pressure Heat Capacity is definied by:   Cp  = ((∂H ̃/∂T)p

Therefore, we obtain the following useful relationships:

                                             Cp = Cv

Remember, Cp = Cv for liquids and solids only for modest changes in T and P.

The trouble with this platform of free discussion is that the people- sometimes but not always- they do not read what the other fellows are writting!!! Especially the case for the present discussion on the Cv  and Cp. 

 The change of entalpy for solid and liquid substances is obtained as a result of a change just in internal energy regardless the change in pressure or volume so Cv = Cp for solid and liquid.. but for gases the enthalpy is a function of temperature and pressure (especially when the change in pressure is higher than few atm.) so the enthalpy is obtained as the change of internal energy plus the effect of volume & pressure changes. for that reason the heat capacity for gases defers when the process is undergoing at P=cost or V=cost . 

Dear Adnan, please look at the steam tables or any other thermophysical data of liquids. Cp is definitely bigger than Cv. They become equal only at the absolute zero temperature. But again, in engineering equipment the actual heat capacity of liquids is somewhere between Cp and Cv.

Btw, it is obviously incorrect to say that liquids are incompressible.

The solids and liquids are considered to be incompressible at a normal working range of temperatures (let's say -30OC to 1000Oc). Hence there is no appreciable change in pressure or volume of the solid and liquids with the application of heat. That gives the same value for CP and CV. 

Prof. Srujan Singh gave complete and detailed answer for this question.  In other words,

If a gas is heated in a sealed container having constant volume it cannot expand, and so no external work is done by the gas during the heating process. All of the thermal energy supplied is used to increase the KE of the molecules or atoms of the gas, and this causes the temperature to rise.

In the case of a gas heated at constant pressure, the gas will expand during the heating process; this is necessary in order to keep the pressure constant despite the rise in temperature. During the expansion, the gas will do work against whatever it is that is maintaining the pressure constant (for example a piston in a cylinder). This external work (We) must be accounted for by the thermal energy provided (Wh), and only the difference (Wh - We) will be available for raising the KE of the gas particles. The temperature will therefore rise less than if the same amount of Wh were absorbed at constant volume.

Specific heat is a measure of the energy required to raise the temperature of 1kg of the material by 1 degree C; this will be greater in the case of heating at constant pressure than it is when heating at constant volume.

In the case of solids and liquids, thermal expansion is much less, so the the amount of external work done when a solid or liquid expands on heating is negligibly small. Consequently the difference between the two specific heats is hardly noticeable, and can be ignored.

Cp and cv exist for compressible media (which we do not neglect the compressibility).

This is usually the case of gas, but it can also be one of liquids and deformable solids.

I totally agree with the detailed analysis of Prof. Maddali V S Murali Krishna. The main reason for your question about the same heat capacity of the solids and liquids is due to the negligible change in the volume during heating of liquids and solids, which in turn results in the same heat capacity value.

@ Artur Benisek

This beautiful relation means that for all systems (gases, liquids, etc) there

is difference between heat capacitance and C_P > C_V. However one must

remember that in the formula derivation the homogeneity of the system is

assumed. For non-homogeneous system (e.g. two-phase coexistence like in the liquid-gas phase transition) it could occur that C_P = C_V.

The only mistake of Srujan Singh is Delta H=Dleta U+Pdelta V+V Delta P

Dear Rasool; you are hundred per cent write in your comment. First of all one defines enthapy as  H=U + PV  that  is called Legendre transformation,  and its purpose of which to get an new characteristic fct in term of pressure not volume as an arguments in addition to the entropy in single component closed systems:  Then takes the differential of which and used the differential form of energy and adds it to  get:

dH=TdS +VdP in closed system, in open multi- component system It takes the following form;

dH= Tds + VdP + Sumk  ChemPotk  Nk  QED.

Note: From the beginning of this discussion I have given full treatment of this problem , which was very trivial in deed for any one who has taken only one semester course on the chemical thermodynamic or physical chemistry  in under graduate level! Good Luck Every Body!!

For the gases, the temperature is increased by either constant volume and constant pressure, known as Cp and Cv. Hence, gases have Cp and Cv. Also in case of solid, the values of Cp and Cv remains almost same, hence solid has only one specific heat.

By definition,The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

To start, imagine a cylinder piston system for constant pressure example. Therefore by first law of thermodynamics (Q=W+ΔU) if you add heat to it, to maintain constant pressure the piston will have to move (work is done). Hence, it will require more heat to raise the temperature since the heat is used to perform work.

Whereas, a constant volume can be imagined as a fully enclosed system. Therefore when heat will be added its sole purpose will be to change the internal energy(Q=ΔU), since no work can be performed (because of enclosed system). Hence, it requires less heat to raise the temperature.

therefore, Cp = Cv+R (skipping the derivation).

Because of this behavior of gases it has two different Specific heats.

For in case of solids, if we heat it it do expand performing little work in expansion (can be considered at constant Atmospheric pressure) giving us a specific heat.

But its not liable to enclose a solid in a constant volume, since it already has a form as compared to formless fluid.

Even it is achieved the difference is not considerable.

So they have only on specific heat.

it means:

if you fixed your volume then you need less heat to achieve 1 temp rise

but

if you fix your pressure you need more heat to achieve 1 temp rise

The so called heat capacities represent the ratio between the energy supplied to a system by means of a thermal interaction (commonly called heating or transfer of heat) and the rise of temperature of the system.

When the system is at constant volume, (assumed that it is isolated from other thermal interactions different that which we are measuring-for example thermal radiation coming from an external source-) the system can not perform work on the surroudings and all the energy supply to it corresponds with the increase of the internal energy of the system.

When the system is at constant pressure, the system can perform work on the surroundings and the supply of energy is shared between the increase of internal energy and the work performed on the surroudings by the expansion of the gas.

So at the same energy supply the increase of internal energy is less when the gas is maintained at constant temperature and when it is maintained at constant pressure.

As for ideal gases the variation of internal energy is proportional to the variation of temperature, a lower increase of internal energy, is externally measured as a lower increase of temperature.

In solids and liquids their dilatation is very much smaller than in gases, so the work performed on the surroundings is small and so it can be neglected and from the experimental point of view both constant and volume specific heat can considered as the same parameter.

In resume the same supply of heat (energy through a thermal interaction) provokes a lower increase of temperature in a system maintained at constant pressure than in the same system maintained at constant pressure. If you want to provoke the same increase of temperature a greater supply of energy (heat) is needed.

Basic thermodynamic laws allow to quantify this difference, in terms of the so called Mayer relation c_p-c_v=R for ideal gases, or generalized Mayer relation (in terms of isobaric dilatation and isothermic compresibility coefficients) for other substances.

We have to remember that to get specific heat, required temperature difference is only one degree, hence with this one degree difference , volume will not change for liquids and solids, so only one specific heat.

For gas we calculate the value of specific heat at constant volume and specific at constant pressure and difference between these two is one degree secondly volume remains constant in case of solid so we consider only one value of specific heat

The general relationship between Cv and Cp for condensed mater as well as for gases under the isothermaL thermodynamic equilibrium conditions (below the critical temperatures and pressures) is given in Guggenheim Thermodynamics book, which is also cited by some authors in this platform in terms of thermal expansion coefficient, compressibility (1/K) and specific molar volume, Vm.

The populer answer Cp= Cv +R relationship appear in this platform is valid only for the perfect gases but nothing else!

For perfect gases as stated also by Guggenheim, where one has following connections:

x=1/k= - d (lnVm)/ dP = P^-1 under constant T

alpha =d Ln ( Vm)/dT =T^-1 under constant P

İf one substitutes above connections in the general formula then obtains above mention populer answer Cp= Cv + R for molar quantities. In the case of van der Waals' gases above relationship should be replaces by

Cp-Cv= R {1+ P R^-1 dB/dT }^2

I agree the answer of Dr. Srujan Singh

I agree also with Dr. Singh, but Dr. Benisek's answer gives the quantitative evaluation of Cp- Cv.

There is a typo in Dr. Benisek's answer for those who are interested, it should read Cp-Cv = TV alpha^2 / K

Thanks!

Specific Heat Capacity is material dependent, from the word specific, it implies the heat capacity per unit mass of a substance. The solid materials have definite shapes, while fluids have varying shapes based on the volume, pressure and temperature as well as the nature of the container they occupy. From the general relationship of PV=mRT, where m=(PV/RT), since the mass, m, depends on Pressure, P, and volume V, then the specific Heat Capacity is also dependent on P and V.

This is why the specific heat of liquids and gases depend on pressure and temperature.

Cp and Cv are equal for incompressible substances. Solids and liquids are incompressible in nature. Hence for solids and liquids Cp and Cv is denoted by the letter ‘C’ only. Cp and Cv concept is for gases only.

This is so simple and can be proved

The fundamental expression below always applies.

Cpm - CVm = T Vm a^2 / kT

In this, the terms are molar heat capacities, temperature, molar volume, isobaric expansivity, and isothermal compressibility.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Physical_Chemistry_(Fleming)/4%3A_Putting_the_First_Law_to_Work/4.2%3A_Compressibility_and_Expansivity

The definition of a perfectly incompressible solid or liquid is when kT = (1/Vm) (dVm/dp)_T = 0. The volume does not change as pressure changes. This leads immediately to the undefined conclusion that Cpm - CVm = infinity for a perfectly incompressible solid???? The specific heats differ by infinity!

The conundrum has only one perfectly realizable solution. Allow that a = 0 as well. In the limit that both a and kT are zero, the undefined ratio of 0^2 / 0 must be resolved with L'Hopital's rule. It becomes T Vm 2a = 0.

Saying that a = 0 is saying that (1/Vm)(dVm/dT)_p = 0. Physically, this means the molar volume (inverse of density) of the solid or liquid does not depend on temperature.

In conclusion, the only way that we find EXACTLY Cpm = CVm for any substance is when the density of that substance does not depend on both pressure AND temperature.

We quote Cpm (or mass specific) values for solids and liquids (and gases) for one simple reason: we live in a constant pressure world. We take CVm ~ Cpm for solids and liquids because we allow that we work mostly over small pressure or temperature ranges so that density (molar volume) does not change. Alternatively, we allow that any change due to increasing pressure (decrease in molar volume) is offset by any change due to increasing temperature (increase in molar volume) OR vica versa. Which ever way you go, you are effectively saying that a = 0 and kT = 0 when you say CVm = Cpm.

We can and do also quote CVm values for gases because the relationship for an ideal gas is easy: Cpm - CVm = R and because we assume most gases are ideal. The fundamental relationship for some forms of real gases can also be solved analytically and/or with approximations.

In summary: We can easily and directly MEASURE Cpm (or its mass specific value) for ALL substances because we live in a constant pressure world. We assume CVm ~ Cpm for "incompressible" substances because we assume the density does not change as we change the pressure and temperature. We can easily DERIVE and therefore quote CVm values for gases from analytical relationships (and/or we measure the values when needed).

I agree with you

This is true because the expansion of gasses is 300 times compared to the solids. So while the process of increasing the temperature by unit degree gas expansion occurs very rapidly and makes the process of heat capacity complicated. So that why to carry out this process either the gasses have to apply contant volume or constant pressure.

liquids and solids cant change their volume with pressure, then the heat needed for change their temperature in one degree centigrade is the same no matter if the process occurs at constant pressure or constant volumecur. As a consequence there are only one value for the specific heat. then the model for solids and liquids consider the same value for the internal energy and the enthalpy.

There is a simpler answer... from the definitions: Cp=dh/dT)p and Cv=du/dT)v also h=u+pv and pv=zRT. If z is small (liquids and solids), the difference between h and u also Cp and Cv is also small.