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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)?

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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
pstnonsonjoku

The pressure that would be applied to steam at 315°c to condense the steam to liquid water is 149 atm.

We have to use the  Clausius-Clapeyron equation to obtain the required pressure;

The equation states that;

ln(P2/P1) = -ΔHvap/R(1/T2 - 1/T1)

Let the reference points P1 and T1 be the pressure and temperature at the boiling point of water.

P1 = 1 atm

T1 = 100°C or 373 K

Now;

T2 = 315°C + 273 = 588 K

R = 8.314 J/mol

ΔHvap = 40.7 kJ/mol or 40700 J/mol

P2 = ?

Substituting values;

ln(P2/1) =- 40700/8.314(1/588 - 1/373)

ln(P2/1) = 5.004

e^ln(P2/1) = e^5.004

P2/1 = e^5.004

P2 = e^5.004 × 1

P2 = 149 atm

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