Liquid sodium can be used as a heat transfer fluid. It's vapor pressure is 40.0 torr @ 633 degrees C and 400.0 torr @ 823 degreees C. Calculate it's heat vaporization.
You want to use the Clausius-Clapeyron equation.
To calculate the heat of vaporization, you can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures to its heat of vaporization. The equation is:
ln(P₂/P₁) = (-ΔH_vap/R) * (1/T₂ - 1/T₁)
Where:
P₁ and P₂ are the initial and final vapor pressures,
T₁ and T₂ are the initial and final temperatures, respectively,
ΔH_vap is the heat of vaporization, and
R is the ideal gas constant (8.314 J/(mol·K)).
However, to use this equation, we need the temperatures in Kelvin. So let's convert them:
T₁ = 633 + 273 = 906 K
T₂ = 823 + 273 = 1096 K
Now, let's substitute the values into the equation:
ln(400.0/40.0) = (-ΔH_vap/8.314) * (1/1096 - 1/906)
Simplifying further:
ln(10) = (-ΔH_vap/8.314) * (0.000915 - 0.001103)
ln(10) = (-ΔH_vap/8.314) * (-0.000188)
Finally, we can solve for the heat of vaporization (ΔH_vap):
ΔH_vap = -8.314 * ln(10) / (-0.000188)
Calculating this expression, we find:
ΔH_vap ≈ 2.06 × 10⁴ J/mol (rounded to two decimal places)
Therefore, the heat of vaporization of liquid sodium is approximately 2.06 × 10⁴ J/mol.
To calculate the heat vaporization of liquid sodium, you need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and heat of vaporization. The equation is given as follows:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 and P2 are the initial and final vapor pressures, respectively,
T1 and T2 are the initial and final temperatures, respectively,
ΔHvap is the heat of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)).
Given:
P1 = 40.0 torr,
T1 = 633 °C = 633 + 273 = 906 K,
P2 = 400.0 torr,
T2 = 823 °C = 823 + 273 = 1096 K.
Now, substitute the given values into the equation and solve for ΔHvap.
ln(400.0/40.0) = (ΔHvap/8.314) * (1/906 - 1/1096)
Simplifying the equation:
ln(10) = (ΔHvap/8.314) * (0.001102 - 0.000912)
ln(10) = (ΔHvap/8.314) * (0.00019)
Now, isolate ΔHvap by multiplying both sides of the equation by (8.314/0.00019):
ΔHvap = ln(10) * (8.314/0.00019)
Using a scientific calculator, calculate the right-hand side of the equation:
ΔHvap ≈ 362,399 J/mol
Therefore, the heat of vaporization of liquid sodium is approximately 362,399 J/mol.