Equation for fluid density

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

vampirchik [111]

Answer:

I) $c=1385.667\frac{J}{kg K}$

II)The difference from the value obtained on part I is: 2000-1385.67 =614.33 $\frac{J}{Kg K}$

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

Explanation:

From the problem we have the molar mass given $M=18\frac{gr}{mol}$ of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.

Part I

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 18.0 g/mol=0.018kg/mol.

Let $C_v (\frac{J}{Kg K})$ the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.

Let $C (\frac{J}{Kg K})$ the specific heat capcity this value represent the heat capacity aboserbed by mass.

For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by $C_v =\frac{R}{2}$ so the total for the 6 degrees of freedom would be:

$C_v =6*\frac{R}{2}=3R=3x8.314\frac{J}{mol K}=24.942\frac{J}{mol K}$

And by definition we know that the specific heat capacity is defined:

$c=\frac{C_V}{M}$

If we replace all the values we have:

$c=\frac{24.942\frac{J}{mol K}}{0.018\frac{kg}{mol}}=1385.667\frac{J}{kg K}$

So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is $c=1385.667\frac{J}{kg K}$

Part II

The actual specific heat of water vapor at low pressures is about 2000 J/(kg * K). Compare this with your calculation.

The difference from the value obtained on part I is: 2000-1385.67 =614.33 $\frac{J}{Kg K}$

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

4 0

3 years ago