In this week\'s experiment, the heat of vaporization of liquid nitrogen is deter

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 69.3oC, 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 41.3oC.

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 -196oC. Based upon your results above, what is the value of Trouton's constant?

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

Explanation / Answer

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

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