I know this question is already on here but the answer wasn't helpful to me.
Liquid Sodium can be used as a heat transfer fluid. Its vapor pressure is 40.0 torr at 633 C and 400.0 torr at 823 C. Calculate its heat of vaporization.
100.016 or 1.00 x 10^2 kj/mol
I looked and did not see another post of this problem; therefore, I don't know what was confusing you. Wouldn't you use the Clausius-Clapeyron equation?
I'm just not sure how to apply it to this problem.
To apply it you simply plug in the numbers, punch them into your calculator, turn the crank and out comes the answers.
ln(p2/p1) = [DHvap/R](1/T1 - 1/T2)
You have p1 @ T1 and p2 @ T2. R is 8.314, solve for DHvap. The unit for DHvap will be joules.
To calculate the heat of vaporization of liquid sodium, you can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at two different temperatures to its heat of vaporization.
The Clausius-Clapeyron 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
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
Let's solve the equation step by step:
Step 1: Convert temperatures to Kelvin
Given:
P1 = 40.0 torr
P2 = 400.0 torr
T1 = 633 °C
T2 = 823 °C
To convert from Celsius to Kelvin, add 273 to the Celsius values:
T1 = 633 + 273 = 906 K
T2 = 823 + 273 = 1096 K
Step 2: Convert pressure to atmospheres
Since the ideal gas constant (R) is typically expressed in J/(mol·K) and atmospheric pressure is often used, we need to convert torr to atmospheres. The conversion factor is 1 atm = 760 torr.
P1 = 40.0 torr / 760 torr/atm = 0.0526 atm
P2 = 400.0 torr / 760 torr/atm = 0.5263 atm
Step 3: Apply the Clausius-Clapeyron equation
Plugging the values into the equation, we have:
ln(0.5263/0.0526) = -ΔHvap/(8.314 J/(mol·K)) * (1/1096 K - 1/906 K)
Step 4: Solve for the heat of vaporization (ΔHvap)
Rearranging the equation to solve for ΔHvap, we have:
-ΔHvap/(8.314 J/(mol·K)) = ln(0.5263/0.0526) / (1/1096 K - 1/906 K)
Now, we can calculate:
-ΔHvap/(8.314 J/(mol·K)) = ln(10) / (1/1096 K - 1/906 K)
Calculating the right-hand side:
ln(10) ≈ 2.3026
1/1096 K - 1/906 K ≈ 0.001214 K^-1
-ΔHvap/(8.314 J/(mol·K)) = 2.3026 / 0.001214
Finally, solve for ΔHvap:
ΔHvap = (-2.3026 / 0.001214) * 8.314 J/mol·K
ΔHvap ≈ -393,847 J/mol·K
Therefore, the heat of vaporization of liquid sodium is approximately -393,847 J/mol·K. The negative sign indicates that heat is being absorbed during vaporization.