Respuesta :
Answer:
Given data:
Equation of the state [tex]p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }[/tex]
Where p = pressure of fluid, pα
T = Temperature of fluid, k
V = Specific volume of fluid [tex]m^{3} / k g[/tex]
R = gas constant , [tex]j/k g k[/tex]
a, b = Constants
Solution:
Specific heat difference, [tex]\begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}[/tex]
According to cyclic reaction
[tex]\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}[/tex]
Hence specific heat difference is
[tex]c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}[/tex]
Equation of state, [tex]p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}[/tex]
Differentiating the equation of state with respect to temperature at constant volume,
[tex]\(\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}\)[/tex]
[tex]\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}[/tex]
Differentiating the equation of the state with respect to volume at constant temperature.
[tex]\(\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)\)[/tex]
Substituting both eq (3) and eq (4) in eq (2)
We get,
[tex]{cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}[/tex]
Specific heat difference equation,
[tex]\(c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}\)[/tex]
