Molar Heat Capacities of Gases, Physics tutorial

Molar Heat Capacities of Gases:

In computing quantity of heat (Q) acquired by the solid or liquid used the expression:

Q = mcΔΘ...............Eq.1

Where, m is mass of the solid/liquid, c is specific heat capacity of substance, and ΔΘ is change in temperature. It is more suitable to use concept of mole to explain amount of the substance particularly gases. By definition, one mole (1 mol) of any substance is quantity of matter such that its mass in grams is numerically equivalent to molecular mass M (frequently known as molecular weight). So to compute number of moles n numerically, divide mass m in grams by molecular mass M.

Therefore n = m/M

Therefore m = n/M............Eq.2

Therefore the Eq.1 becomes

Therefore Q = NMcΔΘ............Eq.3

Q/nΔΘ = Mc............Eq.4

Product Mc is known as the molar heat capacity

C = Mc Q/nΔΘ............Eq.5

This is stated as quantity of heat Q in joules needed to raise temperature of 1 mole of gas through 1 K or 1oC.

Therefore Q = ncΔΘ............Eq.6

Small c is specific heat capacity whereas big C is molar specific heat capacity. It is molar heat capacity which is mostly related with gases although it could also be utilized for solids and liquids. You would have seen that behavior of gases under heat is quite different from behavior of solids and liquids. Molecules of the gas are moving about in different directions with different speeds at any moment. This hence explains random motion of molecules. Internal energy of gas is hence Kinetic Energy (KE) of the random motion. This Kinetic energy depends on temperature of gas. The higher the temperature, the greater the internal energy (kinetic energy) of the gas. Hence, now the ideal gas is stated as one that obeys Boyle's law and whose internal energy relies on temperature of gas and is independent of the volume.

Work Done by the Expanding Gas:

If gas is warmed it expands, gas does external work as it will be seen to have pushed back the piston X against some external pressure P. Heat supplied is shared between work done against external work and in increasing internal energy of the gas as molecules move faster and its Kinetic energy is increased.

Therefore, heat supplied = ΔQ = ΔU +ΔW ..................................Eq.7

Where, ΔQ is increase in heat energy, ΔU is increase in internal energy and ΔW is work done because of expansion, work done against external pressure P.

Expression = ΔQ = ΔU +ΔW is derived from first law of thermodynamics that is also associated to law of conversation of energy. If external pressure is constant with the value of P while volume of gas expands by ΔV, and if area of the piston X is A, moving through the distance d, then increase e in work done against the external pressure

P = ΔW = force x distance ΔW = F x d ......................................... Eq.8

The pressure is expressed as P = AF

Therefore F = PA

Therefore ΔW = PA = PΔV (where, ΔV = Ad)

Therefore ΔQ = ΔU + PΔV

In an ideal gas, work done in separating molecules against attractive forces between them is ignored. This is not so with real gases where Van der Waal's forces have to be considered. It is also supposed that there is no frictional force when piston moves.

Molar Heat Capacities at Constant Volume and Constant Pressure:

Heat capacity of the gas depends on condition under which it is heated that is

  • At constant volume or
  • At constant pressure

At constant volume, we would represent molar heat capacity by CV. At constant pressure, we would represent molar heat capacity by CP.

Molar Heat Capacity at Constant Volume CV:

At constant volume, volume of gas is kept at constant volume in such a way that there is no work done by gas when it absorbs heat. Entire heat is thus utilized in changing internal energy of gas.

Molar heat capacity at constant volume CV, is thus stated as heat needed to raise 1 mole of gas by 1 Kelvin (or 1 Celsius) degree when volume is kept constant.

Cv = Q/nΔΘ

Therefore Q = nCvΔΘ

Where, n is number of moles of gas, CV is molar heat capacity of gas, and ΔΘ is change in temperature of gas. Unit of molar heat capacity at constant volume CV is J mol-1K-1 or J mol-1oC-1. Specific heat capacity at constant volume CV is heat required to raise the temperature of 1kg mass of the gas by 1K or 1oC. Molar mass of hydrogen is 2g thus 1kg = 1000g that is 500 times the mass of 1 mole.

CV = 500 CV for hydrogen Remember unit of CV is J kg-1K-1 or J kg-1oC-1.

At constant volume, thus, all heat supplied to 1 mole of gas is utilized in raising internal energy of gas.

From ΔQ = ΔU + PΔV

As no external work against pressure is done. Hence PΔV is zero. Substitute in above Eq., we get

Therefore ΔQ = ΔU = nCv(T2 - T1) = nCvΔT

Where n = 1mole

Therefore ΔQ = CvΔT

For the ideal gas in which there are no attractive forces among molecules and each molecule has negligible volume CV is independent of volume of gas. If temperature of gas rises from T1 to T2, gain in internal energy (ΔU) is CV (T2 - T1) for one mole of gas no matter what volume gas may be firstly or may lastly occupy. Therefore internal change in energy depends only on temperature change.

Molar Heat Capacity at Constant Pressure CP:

Molar heat capacity of the gas at constant pressure CP is heat needed to raise temperature of 1 mole of gas at constant pressure by 1K or 1oC.

Cp = Q/nΔΘ

Therefore unit of is CP is J mol-1K-1 or J mol-1oC-1

When heat is supplied to change temperature of the gas from T1 to T2, there is change in volume ΔV and increase in internal energy (ΔU). We have known that ΔU, internal energy of the gas, that is independent of volume CV, is for 1K or 1oC change in temperature.

From ΔQ = ΔU + ΔW

Where, ΔW is work done by gas and ΔU = CVΔT for 1 mole of gas.

Therefore ΔQ = Cp = CvΔ + ΔW# for 1 mole of gas

But as we know that ΔW = PΔV

Therefore Cp = CvΔT + PΔV

For the ideal gas from equation of state of the gas, we know that PV = nRT If there is increase in volume of ΔT when there is change in temperature of ΔT, then

Therefore P(V + ΔV) = nR(T + ΔT)

Therefore PV +PΔV = nRT + nRΔT

Therefore PΔV = nRΔT

For 1 mole of gas n = 1

Therefore PΔV = nRΔT

Therefore Cp = CvΔT + RΔV

Where ΔT = 1K or 1oc

Therefore Cp = Cv + R

Therefore Cp-Cv = R

At constant pressure hence, for ideal gas CP is always greater than CV. Difference R is external work done when gas is warmed at constant pressure so that its temperature changes by 1K or1oC.

Isothermal and Adiabatic Expansion of Gases:

When the solid or liquid is heated, you would have seen that volume increase very slightly. Therefore external work done against pressure is extremely small. As a result two molar heat capacities for solids and liquids are practically equal. I.e., specific heat capacities CV and CP are the same. Though, when real gases expand, some work is done against molecular attractive forces. This is internal energy. Van der waal's forces are considerably important in this case.

In CP - CV = R,

R represents external work done when the ideal gas expands at constant pressure. For real gases CP - CV > R by amount of internal work done when gas expands at constant pressure. When the ideal gas is permitted to expand under the constant temperature, procedure is explained as isothermal expansion. Under this state there is no change in internal energy as ΔU depends on ΔT. From First Law of Thermodynamics,


Therefore ΔQ = ΔU + PΔV as ΔU = 0 at isothermal condition

Heat supplied to maintain it at constant temperature then is equivalent to external work done. As temperature is kept constant, the ideal gas thus obeys Boyle's law PV = Constant during isothermal changes.

If the gas, on other hand, is permitted to expand without heat entering or leaving gas (by insulating cylinder and piston) the energy required for external work is taken from internal energy of the gas. This procedure is explained as adiabatic expansion.

From, ΔQ = ΔU + PΔV

Since ΔQ = 0

Therefore PΔV = -ΔU

As a result, temperature of system falls. Contrarily, when gas is compressed under given condition (adiabatic condition) work done on the gas produces a increase in internal energy that is equivalent to work done. This signifies that temperature of gas rises. Gas is then said to have gone through the adiabatic change when no heat enters or leaves system. For adiabatic changes, it can be illustrated that

PVγ = Constant

Where γ = Cp/Cv

TVγ-1 = Constant

Isothermal and adiabatic formulae apply to changes in P, V and T that happen under reversible conditions. The given assumptions are to be noted:

  • No frictional force exists when the piston moves during gas expansion or contraction.
  • No heat is generated in gas by eddies or swirls of gas in expansion or contraction.

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