Problem #4 (10 points) A cartoon of a water molecule appears below. (e) At room

Question

Problem #4 (10 points) A cartoon of a water molecule appears below. (e) At room temperature, six degres of freedom are active in water vapor. Briefly deseribe these degrees of freedom, using words and/or sketches. (b) what is the value ofy = G/C, for water vapor at room temperature? (c) A closed 2.00 liter bottle contains water vapor at a pressure of 1.20 atm. The pressure of the air surrounding the bottle is 1.00 atm. The temperature of both the water vapor and the surrounding air is 295 K. What are the total translational kinetic energy and total rotational kinetic energy of the water molecules inside the bottle? (d) When the bottle is opened, the water vapor expands in order to achieve mechanical equilibrium with the surrouding air. (That is, the water vapor expands in order to match a particular property with that of the surrounding air. What property is this?) What is the temperature of the water vapor at the end of the expansion? (e) How much work did the expanding water vapor do on the surrounding air?

Explanation / Answer

Monoatomic molecule cannot have rotational dof, but as triatomic molecule has a structure if does have rotational dof's.

These six dof are 3 translational along x,y,z axes and three rotational keeping same axes.

Gamma is given by 1+2/dof => 1+2/6 = 1+1/3 = 4/3 = 1.33

Using law of equipartition of energy , total energy will be divided into each dof equally, so

3kT/2 for translational and same to rotational kinetic energy for each molecule

Where k=1.38×10-23 and T=295K.

Pressure of gas remains same.

When bottlw is opened, pressure becomes same i.e 1atm or atmospheric pressure. As surroundings and vapours were at same temperature earlier, they remains same.

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