what pressure would have to be applied to steam at 350.degress C to condense the steam to liquid water (Hvap=40.7kJ/mol)
I got 147,000 torr but I 'm not positive that is correct
To calculate the pressure required to condense steam at a given temperature, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
where:
P1 = initial pressure
P2 = final pressure (what we want to find)
ΔHvap = enthalpy of vaporization (40.7 kJ/mol in this case)
R = gas constant (8.314 J/(mol·K))
T1 = initial temperature (350 °C + 273.15 = 623.15 K)
T2 = final temperature (the boiling point of water at the desired pressure)
First, we need to find the boiling point of water at the given temperature. The boiling point of water at 1 atmosphere (101.3 kPa) is 100 °C, and for every 10 °C increase in temperature, the boiling point roughly doubles. So, we can estimate the boiling point at 350 °C as:
Boiling point ≈ 100 °C * (2 ^ (350 °C / 10 °C))
Boiling point ≈ 100 °C * (2 ^ 35)
Boiling point ≈ 100 °C * 34,359,738.368
Boiling point ≈ 3,435,973,836.8 °C
Next, we substitute the known values into the Clausius-Clapeyron equation:
ln(P2/101.3 kPa) = (-40.7 kJ/mol / (8.314 J/(mol·K))) * (1/3,435,973,836.8 K - 1/623.15 K)
Simplifying the equation gives:
ln(P2/101.3) = (-40.7 / 8.314) * (1/3,435,973,836.8 - 1/623.15)
ln(P2/101.3) = -4.90382 * (0.00000000029082 - 0.001604072)
ln(P2/101.3) = -4.90382 * (-0.001603781928)
ln(P2/101.3) = 0.00784210949
Now, we can solve for P2 by taking the exponent of both sides:
P2/101.3 = e^(0.00784210949)
P2 = 101.3 * e^(0.00784210949)
Calculating this on a scientific calculator or using software, we find that:
P2 ≈ 101.3 * 1.007866999
P2 ≈ 102.075 kPa
Therefore, approximately 102.075 kPa of pressure would need to be applied to condense steam at 350 °C.
To determine the pressure required to condense steam at a particular temperature, we can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at different temperatures to its enthalpy of vaporization.
The Clausius-Clapeyron equation can be written as:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 and P2 are the initial and final pressures respectively,
ΔHvap is the enthalpy of vaporization (40.7 kJ/mol in this case),
R is the gas constant (8.314 J/(mol·K)),
T1 and T2 are the initial and final temperatures respectively.
Given that the steam temperature is 350°C, which is equivalent to 623.15 K, and we want to condense it to liquid water, at which point it would typically be at 100°C, or 373.15 K, we can substitute these values into the equation to find the pressure.
ln(P2/P1) = (40.7 kJ/mol / (8.314 J/(mol·K))) * (1/623.15 K - 1/373.15 K)
Now, let's calculate the natural logarithm:
ln(P2/P1) ≈ 4.897 * (0.0016047 - 0.0026755)
ln(P2/P1) ≈ -0.003924
To solve for P2/P1, we can take the exponent of both sides of the equation:
P2/P1 ≈ e^(-0.003924)
P2/P1 ≈ 0.996086
Finally, to find the pressure we need to apply to the steam, we can rearrange the equation:
P2 = P1 * (P2/P1)
P2 = P1 * 0.996086
Therefore, at a steam temperature of 350°C, the pressure required to condense the steam to liquid water would be approximately 99.6% of the initial pressure.