Question

(a) Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three

translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water

is${\mathbf{M}}_{\mathbf{W}}\mathbf{=}\mathbf{18}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}$(b) The actual specific heat of water vapor at low pressures is about 2000J/kg.k. Compare this with your calculation and comment on the actual role of vibrational motion.

Short answer

The molar heat capacity is equal 24.9 J /mol - K,

The specific heat capacity c of Water vapour at constant volume with 3 transitional and 3 rotational degrees is1380 J/kg . K

there is Increase by 620 J / kg - K due to vibrational motion of molecules

Step-by-step solution

About specific heat capcity

Heat capacity, Cp, is the amount of heat required to change the heat content of 1 mole of material by exactly 1°C.

Heat is a form of energy, often called thermal energy. Energy can be transformed from one form to another (a blender transforms electrical energy into mechanical energy), but it cannot be created nor destroyed; rather, energy is conserved.

Determien the Specific heat at constant volume

We are given a water vapor with molar mass${\mathrm{M}}_{\mathrm{w}}=18\mathrm{g}/\mathrm{mol}$ at constant volume- The vapour molecules have threetransitional and three rotational degrees of freedom.

Solution

a .-

At constant volume means we are about to use molar heat capacity${\mathrm{C}}_{\mathrm{V}}$and before calculating the required let make adifference betWeen ${\mathrm{C}}_{\mathrm{V}}$and c- The uppercase ${\mathrm{C}}_{\mathrm{V}}$is the molar heat capacity where its unit J /mol - K and represent theheat capacity absorbed by mole range. While the lOWercase c is the speci?c heat capacity where its unit J / kg - K andrepresent the heat capacity absorbed by mass range (gram or kilogram) and it is the required to be calculated in part (a).

As given there are 6 degrees of freedom and the molar heat capaczty${\mathrm{C}}_{\mathrm{V}}=\frac{\mathrm{R}}{2}$for each degree (revxew section 18-4 In

textbook), then for the 6 degrees the total molar heat capacity would be

$$\begin{gathered} R \\ {C}_V=6\times \frac{R}{2} \\ =3R \\ =3\times 8.314J/mol·K \\ =24.9J/mol·K \end{gathered}$$

Therefore The molar heat capacity is equal 24.9 J /mol - K,

Determine the specific heat

calculate the specific heat capacity c by known the $\textit{molar heat capacity and molar mass Ml

$$\begin{gathered} c=\frac{C_V}{M} \\ =\frac{24.9J/mol-K}{18\times {10}^{-3}kg/mol} \\ =1380J/kg.K \end{gathered}$$

Therefore The specific heat capacity c of Water vapour at constant volume with 3 transitional and 3 rotational degrees is1380 J/kg . K

Compare the results 

b -

Now compare the result in part (a) with the given value in part (b) we would ?nd that:

For actual vapor water the specific heat capacity c = 2000J/kgK and it is more than the

2000 - 1380 = 620 J/kgK

This increase is due to the Vibrational motion, In part (a) there is no contribution to the vibrational motion so speci?c

heat capacity is lOWer. While in part (b) the given value is for the real vapour water which has additional motion (vibrational

motion) of molecules that required more heat to vibrate and therefore higher speci?c heat capacity

Therefore there is Increase by 620 J / kg - K due to vibrational motion of molecules