Octane has a vapor pressure of 40 torr at 45.1 degrees Celsius and 400 torr at 104 degrees Celsius. What is its heat of vaporization? (Given R=8.314J/mol*K) Please explain.

Why did the octane decide to take a break and go on a hot date with some vapor pressure? Well, it simply wanted to cool down, you know, get a little steamy! But now you're wondering about the heat of vaporization, huh? Don't worry, I've got the funny facts for you!

To figure out the heat of vaporization, we can use the Clausius-Clapeyron equation, which goes like this: ln(P2/P1) = -(ΔHvap/R)((1/T2) - (1/T1)).

Let's plug in the values and get cracking!

First, we'll convert those temperatures to Kelvin, because science likes it that way. So T1 = 45.1 + 273.15 = 318.25 K, and T2 = 104 + 273.15 = 377.15 K.

Now, let's plug in the vapor pressure values. ln(400/40) = -(ΔHvap/8.314)((1/377.15) - (1/318.25)).

Time for some algebraic acrobatics! Simplify the equation. ln(10) = -(ΔHvap/8.314)((1/377.15) - (1/318.25)).

Divide both sides of the equation by -(1/8.314)((1/377.15) - (1/318.25)): ΔHvap = -8.314 * ln(10) * ((1/377.15) - (1/318.25)).

Now grab your calculator and crunch the numbers! ΔHvap ≈ your answer in Joules per mole.

Voila! The heat of vaporization of octane is ready to make its grand entrance, all thanks to some funny equations and the magical powers of math.

To find the heat of vaporization of octane, we can use the Clausius-Clapeyron equation:

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

Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = heat of vaporization
R = ideal gas constant (8.314 J/mol*K)
T1 = initial temperature (in Kelvin)
T2 = final temperature (in Kelvin)

Given:
P1 = 40 torr
P2 = 400 torr
T1 = 45.1 degrees Celsius = 318.25 Kelvin (adding 273.15 to convert to Kelvin)
T2 = 104 degrees Celsius = 377.15 Kelvin

Plugging these values into the equation:

ln(400/40) = (-ΔHvap / 8.314) * (1/377.15 - 1/318.25)

Simplifying further:

ln(10) = (-ΔHvap / 8.314) * (0.0028 - 0.0031)

Taking the natural logarithm of both sides:

2.3026 = -ΔHvap / 8.314 * (-0.0003)

Simplifying:

2.3026 = ΔHvap * (0.028)

Dividing both sides by 0.028:

ΔHvap = 2.3026 / 0.028 ≈ 82.233 J/mol

Therefore, the heat of vaporization of octane is approximately 82.233 J/mol.

To calculate the heat of vaporization, we can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at different temperatures to its heat of vaporization.

The Clausius-Clapeyron equation is given by:
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).
T1 and T2 are the temperatures in Kelvin.

Given data:
P1 = 40 torr
T1 = 45.1°C
P2 = 400 torr
T2 = 104°C
R = 8.314 J/mol*K

First, we need to convert the temperatures from Celsius to Kelvin:
T1 = 45.1 + 273.15 = 318.25 K
T2 = 104 + 273.15 = 377.15 K

Now we can substitute these values into the Clausius-Clapeyron equation:
ln(400/40) = (-ΔHvap/8.314) * (1/377.15 - 1/318.25)

Simplifying the equation, we get:
ln(10) = (-ΔHvap/8.314) * (0.002648 - 0.003145)

Solving for ΔHvap, we rearrange the equation as follows:
ΔHvap = -8.314 * ln(10) * (0.003145 - 0.002648)
ΔHvap = -8.314 * ln(10) * 0.000497

Calculating the value:
ΔHvap ≈ -36.18 J/mol

Therefore, the heat of vaporization of octane is approximately -36.18 J/mol.