In one experiment, the mass of water is 92.6 grams, the initial temperature of the water is 67.6oC, the mass of liquid nitrogen added to the water is 61.4 grams, and the final temperature of the water, after the liquid nitrogen has vaporized, is 35.6oC.

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

To solve this problem, we need to use the equation:

Q = mcΔT

Where:
Q = heat lost or gained
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature

First, let's calculate the heat lost by the warm water:

Q_water = m_water × c_water × ΔT_water

Given:
m_water = mass of water = 92.6 g
c_water = specific heat capacity of water = 4.184 J K-1g-1
ΔT_water = final temperature - initial temperature = (35.6 - 67.6) oC

Substituting the values:

Q_water = 92.6 g × 4.184 J K-1g-1 × (-32.0 oC)

Calculating:

Q_water = -12222.1312 J

Therefore, the heat lost by the warm water is approximately -12222.1312 J.

Now let's calculate the heat of vaporization of nitrogen:

Q_vaporization = m_nitrogen × ΔH_vaporization

Given:
m_nitrogen = mass of nitrogen = 61.4 g

We need to find ΔH_vaporization. To do so, we can use the equation:

ΔH_vaporization = Q_vaporization / m_nitrogen

Substituting the values:

ΔH_vaporization = -12222.1312 J / 61.4 g

Calculating:

ΔH_vaporization = -199.0319 J g-1

Therefore, the heat of vaporization of nitrogen is approximately -199.0319 J g-1.

To calculate the molar heat of vaporization, we need to convert the units from grams to moles. The molar mass of nitrogen (N2) is 28.0134 g/mol.

ΔH_vaporization_molar = ΔH_vaporization / molar_mass_nitrogen

Substituting the values:

ΔH_vaporization_molar = -199.0319 J g-1 / 28.0134 g/mol

Calculating:

ΔH_vaporization_molar = -7.1009 J mol-1

Therefore, the molar heat of vaporization of nitrogen is approximately -7.1009 J mol-1.

Now, let's calculate Trouton's constant. We know that Trouton's constant is the ratio of the enthalpy (heat) of vaporization of a substance to its boiling point (in K).

ΔH_vaporization = Trouton's_constant × boiling_point

Given:
boiling_point_nitrogen = -196 oC = 77 K (convert to Kelvin)

To find the value of Trouton's constant, we can rearrange the equation:

Trouton's_constant = ΔH_vaporization / boiling_point

Substituting the values:

Trouton's_constant = -199.0319 J g-1 / 77 K

Calculating:

Trouton's_constant = -2.5871 J K-1 mol-1

Therefore, the value of Trouton's constant is approximately -2.5871 J K-1 mol-1.