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 · 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