The vapor pressure of nitrogen at several different temperatures is shown below.

Temperature (K) Pressure (Torr)
65 130.5
70 289.5
75 570.8
80 1028
85 1718

1. Use the data to determine the heat of vaporization of nitrogen in kJ/mol.
2.Determine the normal boiling point of nitrogen in Kelvin.

1) 5.92 kj/mol

To determine the heat of vaporization of nitrogen in kJ/mol, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R)(1/T2 - 1/T1)

where P1 and P2 are the pressures at temperatures T1 and T2, ΔHvap is the heat of vaporization, R is the ideal gas constant (8.314 J/(mol·K)), and T1 and T2 are the temperatures in Kelvin.

Let's choose two data points to calculate the heat of vaporization. We'll use T1 = 65 K and T2 = 85 K, corresponding to pressures P1 = 130.5 Torr and P2 = 1718 Torr, respectively.

ln(1718/130.5) = -(ΔHvap/8.314)(1/85 - 1/65)

Simplifying the equation:

ln(13.15) = -(ΔHvap/8.314)(0.01176)

Now, let's rearrange the equation to solve for ΔHvap:

ΔHvap = -8.314 * ln(13.15) / 0.01176

Calculating the value:

ΔHvap ≈ -8.314 * 2.575 / 0.01176 ≈ -1807.3 J/mol

To convert J/mol to kJ/mol, divide the value by 1000:

ΔHvap ≈ -1807.3 J/mol / 1000 ≈ -1.8073 kJ/mol

Therefore, the heat of vaporization of nitrogen is approximately -1.8073 kJ/mol.

To determine the normal boiling point of nitrogen in Kelvin, we can use the data provided. The normal boiling point refers to the temperature at which the vapor pressure of a liquid is equal to the atmospheric pressure, which is generally taken as 760 Torr.

Looking at the data, we can see that the pressure of nitrogen crosses 760 Torr between 75 K and 80 K. To find a more precise value, we can use linear interpolation.

The slope between these two points can be calculated as:

slope = (P2 - P1) / (T2 - T1)
= (1028 - 570.8) / (80 - 75)
= 457.2 / 5
= 91.4 Torr/K

Next, using the point-slope form of the line equation:

P - P1 = slope * (T - T1)

We can substitute one set of data points (P1 = 570.8 Torr, T1 = 75 K) into the equation and solve for T when P = 760 Torr:

760 - 570.8 = 91.4 * (T - 75)

189.2 = 91.4T - 6855
91.4T = 7044.2
T ≈ 77.0 K

Therefore, the normal boiling point of nitrogen is approximately 77.0 K.

To determine the heat of vaporization of nitrogen (ΔHvap) in kJ/mol, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the heat of vaporization, R is the ideal gas constant (8.314 J/(mol·K)), and T1 and T2 are the temperatures in Kelvin.

Let's use the data provided in the table to calculate the heat of vaporization of nitrogen:

Using the data:
T1 = 65 K, P1 = 130.5 Torr
T2 = 85 K, P2 = 1718 Torr

Converting Torr to atm:
P1 = 130.5 Torr * (1 atm/760 Torr) = 0.17171 atm
P2 = 1718 Torr * (1 atm/760 Torr) = 2.2618 atm

Substituting these values into the Clausius-Clapeyron equation:
ln(2.2618 atm/0.17171 atm) = -(ΔHvap/8.314 J/(mol·K)) * (1/85 K - 1/65 K)

Simplifying the equation:
ln(13.1758) = -(ΔHvap/8.314 J/(mol·K)) * (0.01233 K^(-1))

Now, we can rearrange the equation to solve for ΔHvap:
ΔHvap = -8.314 J/(mol·K) * ln(13.1758) / (0.01233 K^(-1))

Converting to kJ/mol:
ΔHvap = -8.314 kJ/(mol·K) * ln(13.1758) / (0.01233 K^(-1))

Calculating this expression will give us the heat of vaporization of nitrogen in kJ/mol.

To determine the normal boiling point of nitrogen in Kelvin, we need to find the temperature at which its vapor pressure reaches 1 atm.

Using the data provided:
P = 1 atm

We can use the Clausius-Clapeyron equation again, this time solving for T2:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Rearranging the equation for T2:
1/T2 = (1/T1) - (ln(P2/P1) / (ΔHvap/R))

Since we want to find the boiling point, we set P2 = 1 atm and solve for T2:

1/T2 = (1/T1) - (ln(1 atm/P1) / (ΔHvap/R))

Thus, T2 = 1 / ((1/T1) - (ln(1 atm/P1) / (ΔHvap/R))).

Substituting the values:
T1 = 65 K

Using the calculated ΔHvap from the previous part, insert the value into the equation provided. Then solve for T2.

Calculating this expression will give us the normal boiling point of nitrogen in Kelvin.

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