Question

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. the actual specific heat of water vapor at low pressures is about 2000 j/(kg * k). compare this with your calculation.

Answer 1

The specific heat at constant volume of water vapor can be calculated by using the equation Cv = (7/2)R, where R is the gas constant and Cv is the molar heat capacity at constant volume. The molar heat capacity of water vapor is calculated to be 28.99 J/mol·K, while the actual specific heat of water vapor at low pressures is about 2000 J/(kg·K). This difference may be attributed to factors not accounted for in our calculation, such as vibrational motion or other interactions.

Step-by-step explanation:

The specific heat at constant volume of water vapor can be calculated using the equation:

Cv = (7/2)R

where R is the gas constant and Cv is the molar heat capacity at constant volume.

Since the water molecule is a nonlinear triatomic molecule, it has three translational and three rotational degrees of freedom. For a monoatomic molecule, Cv would be (3/2)R, and for a linear diatomic molecule, Cv would be (5/2)R.

So, for water vapor, Cv = (7/2)R.

Since the molar mass of water is 18.0 g/mol, we can use this information to calculate the specific heat at constant volume:

Cv = (7/2)R = (7/2)(8.314 J/mol·K) = 28.99 J/mol·K.

The actual specific heat of water vapor at low pressures is about 2000 J/(kg·K).

Comparing this with our calculation, we see that the actual specific heat of water vapor is higher than our calculated value. This could be due to factors such as vibrational motion or other interactions that are not taken into account in our calculation.

Answer 2

The specific heat at constant volume of water vapor based on a triatomic non-linear molecule with six degrees of freedom is calculated to be 1385.67 J/kg·K, using the equipartition theorem.

This theoretical value is lower than the actual 2000 J/(kg·K) for water vapor, which could be due to the exclusion of vibrational contributions or non-ideal interactions.

Step-by-step explanation:

Calculation of Specific Heat at Constant Volume of Water Vapor

To estimate the specific heat at constant volume (Cv) of water vapor based on the assumption of a triatomic non-linear molecule possessing three translational and three rotational degrees of freedom, we use the equipartition theorem.

This theorem states that each degree of freedom contributes ½R (where R is the ideal gas constant) to the molar specific heat. As vibrational motion is not considered contributing in this case, the calculation will involve six degrees of freedom: 3 translational and 3 rotational.

The molar specific heat at constant volume for the molecule is given by:

Cv = degrees of freedom × (½R) = 6 × (½ × 8.314 J/mol·K) = 3 × 8.314 J/mol·K = 24.942 J/mol·K

To convert this to specific heat per kilogram, we use the molar mass of water vapor (18.0 g/mol), converting it to kg/mol for proper units:

specific heat at constant volume (Cv) = 24.942 J/mol·K / (0.018 kg/mol) = 1385.67 J/kg·K

This theoretical value is considerably less than the actual specific heat of water vapor at low pressures, which is about 2000 J/(kg·K). Differences could arise due to factors such as the contribution of vibrational modes at higher temperatures or interactions not considered in the ideal gas approximation.