Ideal Gas Law Chang ein VolumeProblem

[Elzair]

I am having a little trouble on this problem:

For Water at $$25^{\circ}C$$, (Volume Expansion Coefficient) $$\beta = 2.57 * 10^{-4}K^{-1}$$ and (Isothermal Compressibility) $$\kappa = 4.52*10^{-10}Pa^{-1}$$. Suppose you increase the temperature of some water from $$20^{\circ}C$$ to $$30^{\circ}C$$. How much Pressure must you apply to prevent it from expanding?

Can I treat this as an Isobaric expansion and an Isothermal compression?

I think I found the change in Volume like this: $$\frac{\Delta V}{V_{0}}=\beta\Delta T$$

How do I find the Pressure?

 

[tim_lou]

well, I suppose you wish to find ΔP/ΔT, holding V constant,

so, consider P as a function of T and V, we know the partial derivatives relating to V, but not T. So consider, V=const (we are holding V constant), or V(T,P)=const, then

$$dV=\left(\frac{\partial V}{\partial T}\right)_P dT + \left(\frac{\partial V}{\partial P}\right)_T dP=0$$

so that:
$$\left(\frac{\partial P}{\partial T}\right)_V=-\frac{\left(\partial V / \partial T\right)_P}{\left(\partial V/\partial P\right)_T}$$

now, look at the definition of $\beta$ and $\kappa$, what are they (hint: they are related to partial derivatives of V)?

 

[Elzair]

Thanks!

Thanks, I think I got it. I will now post the answer here for others.

$$\frac{dV}{dT} = \frac{\beta}{\kappa}$$

therefore $$\Delta P = \frac{\beta}{\kappa}\Delta T$$

P.S. Does the Tex function on this forum feature a way to represent the therefore symbol (i.e. a triangle of three dots)?