## Related questions with answers

The Tait equation for liquids is written for an isotherm as: $V=V_{0}\left(1-\frac{A P}{B+P}\right)$ where V is molar or specific volume, $V_0$ is the hypothetical molar or specific volume at zero pressure, and A and B are positive constants. Find an expression for the isothermal compressibility consistent with this equation.

Solution

VerifiedIsothermal compressibility is define as

$\kappa=-\dfrac{1}{V}\dfrac{dV}{dP}\bigg|_{T}$

Here we given Tait equation for liquids at isotherm is

$V=V_{0}\left(1-\dfrac{AP}{B+P} \right)$

Where $V$ is molar volume or specific volume, $V_{0}$ is the hypothetical molar or specific volume at zero pressure and $A$ and $B$ are constants. Then the isothermal compressibility is

$\kappa=-\dfrac{1}{V_{0}\left(1-\dfrac{AP}{B+P} \right)}\dfrac{d}{dP}\left( V_{0}\left(1-\dfrac{AP}{B+P} \right) \right)\bigg|_{T}$

Now $V_{0}$ extract from the differential and for second term of differential use differential by part method we get

$=-\dfrac{V_{0}}{V_{0}\left(1-\dfrac{AP}{B+P}\right)}\left[0-\left( \dfrac{A(B+P)-AP}{(B+P)^{2}} \right)\right]$

$=-\dfrac{1}{\left(1-\dfrac{AP}{B+P}\right)}\left[ -\dfrac{AB}{(B+P)^{2}}\right]$

$\kappa=\dfrac{1}{\left(1-\dfrac{AP}{B+P}\right)}\left[ \dfrac{AB}{(B+P)^{2}}\right]$

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