Reduce Melting Temp. by applying Pressure

[Ryomega]

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

\frac{dP}{dT} = \frac{L}{TΔV}

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.

\frac{dP}{dT} = \frac{L}{TΔV}

dP = \frac{L}{TΔV} dT

P = \frac{L log (T)}{ΔV}

L log (T) = PΔV

Log (T) = \frac{PΔV}{L}

T = exp [\frac{PΔV}{L}]

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

 

[rude man]

dp = (L/Δv)dT/T

then how about computing the definite integral instead of the indefinite one?

Δp = ?

 

[Ryomega]

*Face Palm* Set limit from (T) to (T-1) I take it. Thanks

 

[rude man]

*Face Palm* Set limit from (T) to (T-1) I take it. Thanks

Right. And use the appropriate value of T of course.

 

[Ryomega]

Yup yup! Thanks! I feel a tad retarded for not seeing that.

 

[rude man]

I feel a tad retarded for not seeing that.

Don't. It's always a bit rough the first time you run into a new topic ...