Question

In this week's experiment, the heat of vaporization of liquid nitrogen is determined by measuring the change in temperature of a known sample of warm water when liquid nitrogen is added.

In one experiment, the mass of water is 104 grams, the initial temperature of the water is 69.3^oC, the mass of liquid nitrogen added to the water is 60.6 grams, and the final temperature of the water, after the liquid nitrogen has vaporized, is 41.3^oC.

Specific heat of water = 4.184 J K^-1g^-1

How much heat is lost by the warm water?

Heat lost =         J

What is the heat of vaporization of nitrogen in J g^-1?

Heat of vaporization =           J g^-1

What is the molar heat of vaporization of nitrogen?

Molar heat of vaporization =             J mol^-1

Trouton's constant is the ratio of the enthalpy (heat) of vaporization of a substance to its boiling point (in K). The constant is actually equal to the entropy change for the vaporization process and is most often a measure of the entropy in the liquid state. The value of the constant usually lies within the range 70 to 90 J K^-1mol^-1, with a value toward the lower end indicating high entropy in the liquid state.

The normal boiling point of liquid nitrogen is -196^oC. Based upon your results above, what is the value of Trouton's constant?

Trouton's constant =           J K^-1mol^-1

Answer 1

1. Heat lost

Heat Lost = ms(T1-T2) = 104 X 4.184 X (41.3-69.3) = -12.183 KJ

2. What is the heat of vaporization in J g^-1?
A: Just take the first answer and divide it by the mass of the liquid nitrogen

Mass of nitrogen = 60.6 g

heat = -12183/ 60.6 =- 217.54 J / g

3. What is the molar heat of vaporization of nitrogen?
A: Well, we have found the heat of vaporization of nitrogen in J/g, so now we're just multiplying the second answer by 28 (molar mass of N2) = -217.54 X 28 = - 6091.12 J / mole

The normal boiling point of liquid nitrogen is -196C. Based upon your results above, what is the value of Trouton's constant?
A: Trouton's constant (from above) is the heat of vaporization (the molar heat, not the grams) divided by the boiling point. But, Trouton's equation is in units of K (Kelvin), not degrees Celsius, so add 273 to - 196, so the boiling point of Nitrogen is 77K. Therefore, the answer to this question is the answer to the third question,

divided by 77.

- 6091.12 / 77 = 79.10 J/K mol