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Ryomega
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Homework Statement
The molar volume of ice at 273.15K and 101.33 kPa is V1=19.6 cm3
That of water V2=18.00 cm3.
Latent heat of melting of ice is L=6.0 kJ mol-1.
Find the pressure that must be applied to reduce the melting point by 1 K.
Homework Equations
Clausius-Clapeyron
[itex]\frac{dP}{dT}[/itex] = [itex]\frac{L}{TΔV}[/itex]
ΔV = (V2-V1)
The Attempt at a Solution
The problem I am having is fundamental, this is not the usual find vapour pressure question, which is what caught me off guard. So what I'm asking is a way to set this question up. This is where I've gotten so far.
[itex]\frac{dP}{dT}[/itex] = [itex]\frac{L}{TΔV}[/itex]
dP = [itex]\frac{L}{TΔV}[/itex] dT
P = [itex]\frac{L log (T)}{ΔV}[/itex]
L log (T) = PΔV
Log (T) = [itex]\frac{PΔV}{L}[/itex]
T = exp [[itex]\frac{PΔV}{L}[/itex]]
Do I subtract 1 from T and solve for P? It doesn't feel right. Or did I completely mess it up?
If so please show me the set up, I should be ok with the rest.
Thanks in advance