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ChemistryEdu Logo Thermodynamics | Heat Capacity#

Heat Capacity#

  • We can define heat capacity as the amount of heat required to raise the temperature of a given mass of substance by 1 Kelvin (or 1 ℃).
  • It is denoted by C and is an extensive property, ie, it depends on the amount of matter present in the substance.
\[C = {dq \over dT}\]
\[where,\ q = heat\ and\ T = temperature\]

Specific Heat Capacity#

  • Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 K.
  • It can also be defined as the heat capacity per unit gram of mass.
  • It is an intensive property because it does not depend on amount of matter present in the substance (it is expressed as per unit gram of mass).
  • It is denoted by the symbol S.
\[S = {C \over m}\]
\[where,\ C = Heat\ capacity\ and\ m = mass\ in\ gram\]
\[S = {dq \over mdT}\]

Molar Heat Capacity#

  • Molar heat capacity is defined as the amount of heat required to raise the temperature of 1 mole of a substance by 1 K.
  • It is defined as heat capacity per unit mole.
  • It is an intensive property, ie it does not depend on amount of matter present in the substance.
  • It is denoted by Cm.
\[C_m = {C \over n}\]
\[where,\ C = heat\ capacity\ and\ n = moles\]
\[C_m = {dq \over ndT}\]
\[Also,\ dq = nC_mdT\]

Let's put substitute values in the equation of Cm.

\[C_m = {dq \over ndT}\]

Using first law of thermodynamics, dq = dU - dW:

\[C_m = {(dU - dW) \over ndT}\]
\[C_m = {dU \over ndT} - {dW \over ndT}\]
\[C_m = {nC_vdT \over ndT} - {(-PdV) \over ndT}\]
\[C_m = C_v + {PdV \over ndT}\]
  • Molar heat capacity at constant volume is denoted by Cv.
  • Molar heat capacity at constant pressure is denoted by Cp.

Relation between Cp and Cv#

  • The relation between Cp and Cv is given as:
\[C_p - C_v = R\]

  • Let's try to prove this relation.

  • At constant pressure:

\[dH = dU + PdV\]
\[dH = dU - dW\]
\[dU = dH + dW\]
\[By\ first\ law\ of\ thermodynamics:\]
\[dU = dq + dW\]
\[Hence,\ at\ constant\ pressure,\ we\ can\ write:\]
\[dq = dH\]
\[dq = nC_pdT\ (Since,\ dH = nC_pdT)\]
  • Also, we know that:
\[dU = nC_vdT\]
  • Applying first law of thermodynamics at constant pressure:
\[dU = dq + dW\]
\[nC_vdT = nC_pdT - PdV\]
\[n(C_v - C_p) dT = -PdV\]
\[n(C_v - C_p) dT = -nRdT\]
\[C_v - C_p = -R\]
\[C_p - C_v = R\]

Poisson's Ratio#

Poisson's ratio is defined as the ratio of molar heat capacity at constant pressure (Cp) to molar heat capacity at constant volume (Cv).

\[γ = {C_p \over C_v}\]

Calculation of molar heat capacity and Poisson's ratio for a mixture of gases#

  • Let us consider a container containing a mixture of three gases A, B and C whose moles are n1, n2 and n3 respectively. Let their molar heat capacity at constant volume be Cv1, Cv2 and Cv3 respectively.

  • We can calculate Cv(mix) of mixture of these gases as:

\[C_{v(mix)} = {n_1C_{v1} + n_2C_{v2} + n_3C_{v3} \over n_1 + n_2 + n_3}\]
  • Cp(mix) can be calculated as:
\[C_{p(mix)} = C_{v(mix)} + R\]
  • Poisson's ratio, γmix is given by:
\[γ_{mix} = {C_{p(mix)} \over C_{v(mix)}}\]

Important formula to remember from this article

  • dU = nCvdT

  • dq = nCpdT

  • Cp - Cv = R

  • γ = Cp / Cv

Here, U = internal energy, Cv = molar heat capacity at constant volume, Cp = molar heat capacity at constant pressure, n = moles, q = heat, T = temperature, γ = Poisson's ratio