Question 6.6: To establish the positivity of the specific heat at constant...

To establish the positivity of the specific heat at constant pressure C_p compressibility coefficient at constant temperature k_T  (see relations (6.31)), follow the steps given below:

C_p ≥0        and         κ_T ≥ 0             (6.31)

a) Show that the Mayer relation (5.42) can be recast as,

C_p (T, p) = C_V (T, V) + T \frac{∂p (T,V)}{∂T} \frac{∂V (T,p)}{∂T} .                       (5.42)

C_p = C_V + \frac{α^2}{κ_T} VT .

where α is the thermal coefficient of expansion,

α = \frac{1}{V} \frac{∂V (T,p)}{∂T}       and      κ_T = -\frac{1}{V} \frac{∂V (T,p)}{∂p} .

b) Show that

\frac{\partial^2 F (T,V)}{\partial V^2} = \frac{\frac{\partial^2 U }{\partial S^2 }\frac{\partial^2 U}{\partial V^2} - \Bigl(\frac{\partial^2 U }{\partial S \partial V }\Bigr)^2 }{\frac{\partial^2 U}{\partial S^2} } .

c) Conclude from these two results that κ_T ≥ 0 and C_p ≥ 0.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

a) The Mayer relation yields,

C_p = C_V + T \frac{\partial p}{\partial T} \frac{\partial V}{\partial T} .

According to the cyclic rule,

\frac{\partial p}{\partial T} \frac{\partial T}{\partial V} \frac{\partial V}{\partial p} = -1      thus    \frac{\partial p}{\partial T} = – \frac{\partial V}{\partial T} \frac{\partial p}{\partial V} = \frac{α}{κ_T} .

Moreover,

\frac{\partial V}{\partial T} = α V .

Thus, the Mayer relation can be recast as,

C_p = C_V + \frac{α^2}{κ_T} V T .

b) Using the mathematical definition (4.77),

\frac{\partial f}{\partial y}\mid _z ≡ \frac{\partial f \Bigl(x(y,z),y\Bigr) }{\partial x (y,z)} \frac{\partial x (y,z)}{\partial y } + \frac{\partial f \Bigl(x(y,z),y\Bigr) }{\partial y } .

\frac{\partial f}{\partial z}\mid _y ≡ \frac{\partial f \Bigl(x(y,z),y\Bigr) }{\partial x (y,z)} \frac{\partial x (y,z)}{\partial z } .                      (4.77)

\frac{∂^2F (T, V)}{∂V^2} = \frac{d}{dV} \Biggl(\frac{dF\Bigl(T(S,V),V\Bigr) }{dV}\Biggr) = – \frac{dp\Bigl(T(S,V),V\Bigr) }{dV}

= – \frac{∂p}{∂T} \frac{∂T}{∂V}-\frac{∂p}{∂V} = – \frac{∂p}{∂T} \frac{∂}{∂V} \Bigl(\frac{∂U}{∂S}\Bigr) + \frac{∂}{∂V} \Bigl(\frac{∂U}{∂V}\Bigr) .

Thus,

\frac{∂^2F (T, V)}{∂V^2} = \frac{∂^2U}{∂V^2} – \frac{∂p}{∂T} \frac{∂^2U}{∂S ∂V } .

According to the cyclic rule,

\frac{\partial p}{\partial T} \frac{\partial T}{\partial S} \frac{\partial S}{\partial p} = -1      thus    \frac{\partial p}{\partial T} = – \frac{\partial p}{\partial S} \frac{\partial S}{\partial T} .

which is then recast as,

\frac{\partial p}{\partial T} = – \frac{\frac{∂p}{∂S }}{\frac{∂T}{∂S  } } = \frac{\frac{\partial }{\partial S}\Bigl(\frac{\partial U}{\partial V} \Bigr) }{\frac{\partial }{\partial S} \Bigl(\frac{\partial U}{\partial S} \Bigr)} = \frac{\frac{\partial^2 U}{\partial S \partial V} }{\frac{\partial^2 U}{\partial S^2} } .

Hence,

\frac{\partial^2 F(T,V) }{\partial V^2 }= \frac{\frac{\partial^2 U }{\partial S^2 }\frac{\partial^2 U}{\partial V^2} – \Bigl(\frac{\partial^2 U}{\partial S \partial V } \Bigr)^2}{\frac{\partial^2 U}{\partial S^2} } .

c) According to relation (6.32), the compressibility coefficient κ_T is expressed as,

\frac{\partial^2 F}{\partial T^2} = -\frac{\partial S}{\partial T} = – \frac{C_V}{T} \leq 0           and        \frac{\partial^2 F}{\partial V^2} = -\frac{\partial p}{\partial V} = \frac{1}{κ_T V} ≥ 0 .          (6.32)

κ_T = V \Bigl(\frac{\partial^2 F}{\partial V^2} \Bigr)^{-1} .

According to relation (6.22), the numerator in the expression for ∂^2F (T, V)/∂V^2 is positive and according to relation (6.14), the denominator is also positive. Thus, since the volume V is positive, the compressibility coefficient κ_T is positive, i.e. κ_T ≥ 0. Furthermore, since the temperature is positive, the Mayer relation requires that C_p ≥ C_V. Since C_V is positive according to relation (6.29), this in turn implies that C_p is positive, i.e. C_p ≥ 0.

\frac{\partial^2 U}{\partial S^2} \frac{\partial^2 U}{\partial V^2} – \Bigl(\frac{\partial^2 U}{\partial S ∂V} \Bigr)^2 ≥0 .              (6.22),

\frac{\partial^2 U}{\partial S^2}  ≥0 .        (6.14)

\frac{\partial^2 U}{\partial S^2} = \frac{\partial T}{\partial S}  = \frac{ T}{C_V}≥0                         and            \frac{\partial^2 U}{\partial V^2} = -\frac{\partial p}{\partial V}  = \frac{ 1}{k_S V}≥0 .                          (6.29)

Related Answered Questions