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

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

Use the data to determine the heat of vaporization of nitrogen.

Determine the normal boiling point of nitrogen.

5.912

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

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Where:
P1 = pressure at the first temperature (65°C in this case)
P2 = pressure at the second temperature (boiling point of nitrogen)
T1 = temperature at the first pressure (65°C in this case)
T2 = temperature at the second pressure (unknown in this case)
ΔHvap = heat of vaporization of nitrogen
R = gas constant (8.314 J/mol·K)

Now let's solve for the heat of vaporization (ΔHvap):

We'll use the data at 65°C (338 K) and the boiling point of nitrogen.

For T2, we can assume the boiling point of nitrogen at normal atmospheric pressure, 1 atm, which is approximately -196°C (-321°F) or 77 K.

Using the equation with these values:

ln(P2/130.5) = -ΔHvap/(8.314 J/mol·K) * (1/77 - 1/338)

Simplifying the equation:

ln(P2/130.5) = -ΔHvap/(8.314 J/mol·K) * (-261/26,026)

Taking the exponential of both sides:

P2/130.5 = e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))

Rearranging the equation:

P2 = 130.5 * e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))

Now, let's plug in the values and solve for ΔHvap:

P2 = 130.5 * e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))
289.5 = 130.5 * e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))
570.8 = 130.5 * e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))
1028 = 130.5 * e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))
1718 = 130.5 * e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026))

Before solving for ΔHvap, we can ignore the constant (130.5) for now. When we take the ratios of subsequent equations, this constant will cancel out.

Now, let's take the ratio of subsequent equations to eliminate the constant:

(e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026)) / e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026)) = 570.8/289.5

This simplifies to:

e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026)) / e^(-ΔHvap/(8.314 J/mol·K) * (-261/26,026)) = 570.8/289.5

Since the exponents are equal, we can equate them:

-ΔHvap/(8.314 J/mol·K) * (-261/26,026) = ln(570.8/289.5)

Now, let's solve for ΔHvap:

-ΔHvap/(8.314 J/mol·K) * (-261/26,026) = ln(570.8/289.5)
ΔHvap = [ln(570.8/289.5)] * (8.314 J/mol·K) * (26,026/261)

Calculate this expression to find the heat of vaporization of nitrogen.

To determine the heat of vaporization of nitrogen, we can use the Clausius-Clapeyron equation. This equation relates the heat of vaporization to the vapor pressure and temperature. The equation is as follows:

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.
T1 and T2 are the temperatures at which the vapor pressures are measured.

We have the pressure values and the corresponding temperatures for nitrogen:

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

To use the Clausius-Clapeyron equation, we need to select two data points. Let's choose the points (70, 289.5) and (80, 1028). We can plug these values into the equation:

ln(1028/289.5) = (-ΔHvap/R) * (1/80 - 1/70)

Now we can solve for ΔHvap. Rearranging the equation:

ΔHvap = -(ln(1028/289.5) / (1/80 - 1/70)) * R

R is the ideal gas constant, which is equal to 8.314 J/(mol·K).

Now, substituting the values:

ΔHvap = -(ln(1028/289.5) / (1/80 - 1/70)) * 8.314 J/(mol·K)

Evaluating this expression will give us the heat of vaporization of nitrogen.

To determine the normal boiling point of nitrogen, we need to examine the vapor pressure-temperature relationship. The normal boiling point is the temperature at which the vapor pressure equals the atmospheric pressure, which is approximately 1 atm (760 mmHg).

Looking at the data provided, we can see that the vapor pressure is below 760 mmHg at all measured temperatures. Therefore, the normal boiling point of nitrogen is below the lowest temperature measured (65°C or 338K).

Use the Clausius-Clapeyron equation. A P1 and P2 together with T1 and T2 will get delta Hvap. I might do two or three to see how they agree and take an average. After delta Hvap is determined, you can go to one of the P values (I would choose the pressure closest to the boiling point of N2--you can look that up on the web) and the T for one set of data, then use 760 mm P for the other pressure point and T that corresponds to that will be the boiling point. (The idea here is that the boiling point of a liquid is defined as the temperature at which the vapor pressure of the liquid equals atmospheric pressure. (Remember that's why water boils below 100 C at the top of Pike's Peak. Don't forget to use T in Kelvin.