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The Tait equation for liquids is written for an isotherm as: V=V0(1APB+P)V=V_{0}\left(1-\frac{A P}{B+P}\right) where V is molar or specific volume, V0V_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.

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Answered 2 years ago
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Isothermal compressibility is define as

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

Here we given Tait equation for liquids at isotherm is

V=V0(1APB+P)V=V_{0}\left(1-\dfrac{AP}{B+P} \right)

Where VV is molar volume or specific volume, V0V_{0} is the hypothetical molar or specific volume at zero pressure and AA and BB are constants. Then the isothermal compressibility is

κ=1V0(1APB+P)ddP(V0(1APB+P))T\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 V0V_{0} extract from the differential and for second term of differential use differential by part method we get

=V0V0(1APB+P)[0(A(B+P)AP(B+P)2)]=-\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]

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

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

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