Question
Download Solution PDFFor a compressible system, V, S, P and T are specific volume, specific entropy, pressure and temperature respectively. As per Maxwell's relation
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
If there is a relation between x, y & z, then z may be expressed as a function of x & y.
An equation of type dz = Mdx + Ndy is exact differentials, it can be written as \({\left( {\frac{{\partial M}}{{\partial y}}} \right)_x} = {\left( {\frac{{\partial N}}{{\partial x}}} \right)_y}\)
For a pure substance undergoing a reversible process, these equations hold true:
- dU = T.ds – p.dv
- dH = T.ds + v.dp
- dF = -p.dV – s.dT
- dG = v.dP –s.dT
They can be written as
- \({\left( {\frac{{\partial T}}{{\partial v}}} \right)_s} = - {\left( {\frac{{\partial p}}{{\partial s}}} \right)_v}\)
- \({\left( {\frac{{\partial T}}{{\partial p}}} \right)_s} = \;{\left( {\frac{{\partial v}}{{\partial s}}} \right)_p}\)
- \({\left( {\frac{{\partial p}}{{\partial T}}} \right)_v} = \;{\left( {\frac{{\partial s}}{{\partial v}}} \right)_T}\)
- \({\left( {\frac{{\partial v}}{{\partial T}}} \right)_p} = - {\left( {\frac{{\partial s}}{{\partial p}}} \right)_T}\)
These are known as Maxwell’s Equation.
Out of these options, 4 does not match with Maxwell’s Equation.
Other Maxwell relations are:
if x = ϕ(y, z)
\(dx = {\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}dy + {\left( {\frac{{\partial x}}{{\partial z}}} \right)_y}dz\)
\(or,\;{\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}{\left( {\frac{{\partial z}}{{\partial x}}} \right)_y}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - 1\)
If F = ϕ(x, y, z)
\({\left( {\frac{{\partial x}}{{\partial y}}} \right)_F}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_F}{\left( {\frac{{\partial z}}{{\partial x}}} \right)_F} = 1\)
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