if a pure liquid in a sealed container is in equilibrium with its own vapor, the dependence of the vapor pressure on the temperature and the heat of vaporization is governed by the Clausius-Clapeyron equation.
what is the heat of vaporization if the vapor pressure of liquid water at 25 °C is 23.8 Torr and at a boiling point of 100 °C is 760 Torr.
To find the heat of vaporization, we can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at two different temperatures to the heat of vaporization.
The Clausius-Clapeyron equation is given as:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the vapor pressures at two different temperatures (in this case, 25 °C and 100 °C).
ΔHvap is the heat of vaporization (what we need to find).
R is the gas constant (8.314 J/(mol·K)).
T1 and T2 are the corresponding temperatures in Kelvin (298 K and 373 K).
Now, let's plug in the values we have:
ln(760/23.8) = (-ΔHvap/8.314) * (1/373 - 1/298)
First, let's calculate the natural logarithm (ln) of the ratio of the vapor pressures:
ln(760/23.8) ≈ 5.819
Next, let's simplify the equation:
5.819 = (-ΔHvap/8.314) * (0.002674 - 0.003356)
Now, let's calculate the difference in the reciprocal of the temperatures:
0.002674 - 0.003356 ≈ -0.000682
5.819 = (-ΔHvap/8.314) * (-0.000682)
To find the heat of vaporization (ΔHvap), rearrange the equation and solve:
ΔHvap = (5.819 * 8.314) / 0.000682
ΔHvap ≈ 71,036 J/mol
Therefore, the heat of vaporization of liquid water is approximately 71,036 J/mol.
To find the heat of vaporization using the Clausius-Clapeyron equation, we need to know the vapor pressure of the liquid at two different temperatures. Let's denote the vapor pressure at the lower temperature (25 °C) as P1 (23.8 Torr) and the vapor pressure at the higher temperature (100 °C, the boiling point of water) as P2 (760 Torr). We'll also denote the absolute temperatures as T1 (25 °C + 273.15) and T2 (100 °C + 273.15).
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -(ΔHvap/R)((1/T2) - (1/T1))
Where:
- P2 and P1 are the vapor pressures at the higher and lower temperatures, respectively.
- T2 and T1 are the absolute temperatures at the higher and lower temperatures, respectively.
- ΔHvap is the heat of vaporization.
- R is the ideal gas constant (8.314 J/(mol·K)).
To simplify the equation, we can convert the pressures from Torr to atm by dividing them by 760. Substituting the known values into the equation, we get:
ln(760/23.8) = -(ΔHvap/8.314)((1/(100 + 273.15)) - (1/(25 + 273.15)))
Using the natural logarithm of the ratio of pressures, we get:
6.132 = -ΔHvap/8.314((1/373.15) - (1/298.15))
Now, we can solve for ΔHvap:
ΔHvap = -6.132 × 8.314((1/373.15) - (1/298.15))
Calculating this expression gives us the value of the heat of vaporization for water.