What pressure would have to be applied to steam at 315°c to condense the steam to liquid water (δh vap = 40.7 kj/mol)?

1 answer:

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1 answer · Chemistry

Best Answer

Water steam condenses if its pressure is equal to vapor saturation vapor pressure.

Use the Clausius-Clapeyron relation.
I states the temperature gradient of the saturation pressure is equal to the quotient of molar enthalpy of phase change divided by molar volume change due to phase transition time temperature:
dp/dT = ΔH / (T·ΔV)
Because liquid volume is small compared to vapor volume
ΔV in vaporization is approximately equal to to the vapor volume. Further assume ideal gas phase:
ΔV ≈ V_v = R·T/p
Hence
dp/dT = ΔHv / (R·T²/p)
<=>
dlnp/dT = ΔHv / (R·T²)

If you solve this DE an apply boundary condition p(T₀)= p₀.
you get the common form:
ln(p/p₀) = (ΔHv/R)·(1/T₀ - 1/T)
<=>
p = p₀·exp{(ΔHv/R)·(1/T₀ - 1/T)}

For this problem use normal boiling point of water as reference point:
T₀ =100°C = 373.15K and p₀ = 1atm
Therefore the saturation vapor pressure at
T = 350°C = 623.15K
is
p = 1atm ·exp{(40700J / 8.314472kJ/mol)·(1/373.15K - 1/623.15K)} = 193 atm
hope this helps

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$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

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(d)Mass of O

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3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$