Suppose we make a series of measurements in the laboratory and think we discovered a new thermodynamic property, call it [latex]z[/latex]. Our experimental data provide an empirical equation of the form: [latex]dz = p dv + v^2 dp[/latex]. Is [latex]z[/latex] a new property?

The unknown is whether or not [latex]z[/latex] is a new thermodynamic property. Equation (11.11) here has the form [latex]dz=(\frac{\partial z}{\partial x} )_y dx+(\frac{\partial z}{\partial y} )_x dy=Mdx+Ndy[/latex] (11.11) [latex]dz = Mdx+ Ndy = p dv +v^2dp[/latex] so [latex]M = p, N = v^2 , x = v[/latex], and [latex]\text{y} = p[/latex]. The cross differentials in Eq. (11.12) are [latex](\frac{\partial M}{\partial y} )_x\equiv (\frac{\partial N}{\partial x} )_y[/latex] (11.12) [latex](\frac{∂M}{∂\text{y}} )_x = (\frac{∂p}{∂p} )_v = 1[/latex] and [latex](\frac{∂N}{∂x} )_\text{y} = (\frac{∂ (v)^2}{∂v} )_p = 2v ≠ (\frac{∂M}{∂\text{y}} )_x[/latex] Since Eq. (11.12) is not satisfied here, then z cannot be a thermodynamic property.

Question 11.4: Suppose we make a series of measurements in the laboratory a...

Modern Engineering Thermodynamics

Suppose we make a series of measurements in the laboratory and think we discovered a new thermodynamic property, call it $z$ . Our experimental data provide an empirical equation of the form: $dz = p dv + v^2 dp$ . Is $z$ a new property?

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The unknown is whether or not $z$ is a new thermodynamic property. Equation (11.11) here has the form

$dz=(\frac{\partial z}{\partial x} )_y dx+(\frac{\partial z}{\partial y} )_x dy=Mdx+Ndy$ (11.11)

$dz = Mdx+ Ndy = p dv +v^2dp$

so $M = p, N = v^2 , x = v$ , and $\text{y} = p$ . The cross differentials in Eq. (11.12) are

$(\frac{\partial M}{\partial y} )_x\equiv (\frac{\partial N}{\partial x} )_y$ (11.12)

$(\frac{∂M}{∂\text{y}} )_x = (\frac{∂p}{∂p} )_v = 1$

and

$(\frac{∂N}{∂x} )_\text{y} = (\frac{∂ (v)^2}{∂v} )_p = 2v ≠ (\frac{∂M}{∂\text{y}} )_x$

Since Eq. (11.12) is not satisfied here, then z cannot be a thermodynamic property.