Question

How do I calculate the new pressure of liquid water using its isothermal compressibility $\kappa $and expansion coefficient $\alpha $

Answer 1

Hint: In this question, we need to find the new pressure of liquid water using its isothermal compressibility $\kappa $and expansion coefficient $\alpha $. Isothermal compressibility $\kappa $ helps to determine the compressibility properties of the reservoir. The Expansion coefficient $\alpha $is also known as thermal expansion, The volume of a material increases with an increase in temperature.

Complete answer:
Isothermal compressibility $\kappa $ measures the fractional change in volume with a change in pressure.
$\kappa =\,\dfrac{{{(\dfrac{Partial\,V}{\partial P})}_{T}}}{V}$= $-\dfrac{1}{V}{{(\dfrac{\partial V}{\partial P})}_{T}}$ In this ($\dfrac{\partial V}{\partial P}{{)}_{T}}$ is considered as the partial derivative of volume w.r.t Pressure and the temperature is constant.
$\kappa \partial {{P}_{T}}=\,-\dfrac{1}{V}\partial {{V}_{T}}$
Expansion coefficient $\alpha $is the change in the volume of the substance as soon as there is an increment in the temperature.
$\alpha \,=\,\dfrac{{{(\dfrac{Partial\,V}{\partial T})}_{P}}}{V}$= $\dfrac{1}{V}{{(\dfrac{\partial V}{\partial T})}_{P}}$ In this ${{(\dfrac{\partial V}{\partial T})}_{P}}$is considered as the partial derivative of volume w.r.t Temperature and pressure is constant.
$\alpha \,\partial {{T}_{P}}=\,\dfrac{1}{V}\partial {{V}_{P}}$
Derivative of the differential equation is
$dV=\,{{(\dfrac{\partial V}{\partial T})}_{P}}dT\,+\,{{(\dfrac{\partial V}{\partial T})}_{T}}dP$
Dividing the equation by partial differential temperature where the volume is constant
${{(\dfrac{\partial P}{\partial T})}_{V}}\,=\,\dfrac{-{{(\dfrac{\partial V}{\partial T})}_{P}}}{{{(\dfrac{\partial V}{\partial P})}_{T}}}$
Substituting this with$\kappa $and $\alpha $
${{(\dfrac{\partial P}{\partial T})}_{V}}=\,\dfrac{\alpha }{\kappa }$
By Partial differentiation, we get
${{P}_{2}}-{{P}_{1\,}}=\,\dfrac{\alpha }{\kappa }({{T}_{2}}-{{T}_{1}})$
The final equation would be
${{P}_{2}}_{\,}=\,\dfrac{\alpha }{\kappa }({{T}_{2}}-{{T}_{1}})\,+\,{{P}_{1}}$
So, Isothermal compressibility $\kappa $of water is $4.7\,\times \,{{10}^{-5}}\,at{{m}^{-1}}$
And Expansion coefficient $\alpha $of water is $1.7\,\times \,{{10}^{-4}}\,{{k}^{-1}}$
${{P}_{2\,}}=\,1atm\,+\,\dfrac{1.7\,\times \,{{10}^{-4}}\,{{K}^{-1}}}{4.7\times {{10}^{-5\,}}at{{m}^{-1}}}(6k)$
     = 23 atm
So, the new pressure of liquid water using its isothermal compressibility $\kappa $ and expansion coefficient $\alpha $ is 23 atm.

Note:
The compressibility factors have many applications like It forms a liquid when we compress petroleum gas. In oxygen cylinders that are used for medical facilities. It forms CNG. Methane gets compressed and used as fuel in vehicles.