Question

proof of mayer's relation

Answer 1

Proofs given can range in length to a surprising degree, according to the assumed level of familiarity with staple formulas. One of the shortest I have seen is this one from E. Brian Smith’s Basic Chemical Thermodynamics:

H = U + PV
(q)p = (q)v + PV
For one mole of a perfect gas PV = RT and thus
(q)p = (q)v + RT
Differentiating with respect to temperature and noting that C = (dq/dT) we obtain
Cp – Cv = R

At the other end of the scale is this one, given by Enrico Fermi during a course of lectures at Columbia University in 1936. Quite a masterclass this – he builds the proof from first principles and doesn’t cut any corners whatsoever in his argumentation. I have tried to preserve the nomenclature he used:

The statement of the first law for the system under consideration is

(1)

If we choose T and V as our independent variables, U becomes a function of these variables so that:

(2)

and the first law statement becomes:

(3)

Similarly, taking T and p as independent variables, we have:

(4)

We define the thermal capacity of a body as the ratio dQ/dT. Let Cv and Cp be the thermal capacities at constant volume and at constant pressure, respectively. A simple expression for Cv can be obtained from (3). For an infinitesimal transformation at constant volume, dV=0; hence,

(5)

Similarly, using (4) we obtain the following expression for Cp

(6)

Fermi comments here: “The second term on the right represents the effect on the thermal capacity of the work performed during the expansion. An analogous term is not present in (5), because in that case the volume is kept constant so that no expansion occurs.” This is exactly Mayer’s hypothesis*.

Fermi then asserts on the basis of experimental evidence (namely Joule’s), but without theoretical proof (he does that later), that for an ideal gas U = U(T).

Since U depends only on T, it is not necessary to specify that the volume is to be kept constant in the derivative (5); so that for an ideal gas we may write:

(7)

For an ideal gas, (1) takes on the form

(8)

Differentiating the equation of state for one mole of an ideal gas, pV = RT, we obtain:

(9)

Substituting this for pdV in (8), we find:

(10)

Since dp=0 for a transformation at constant pressure, this equation gives us:

(11)

Which is Mayer’s relation.

*the second term on the right equates to Cp–Cv since for an ideal gas (∂U/∂T)P = dU/dT = CV

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Answer 2

Solution:

$\underline{\bf Mayer's\;relation:-}\\ \\ \implies \sf C_{p}-C_{v}=R\\ \\ \\ \underline{\bf Proof:-}\\ \\ \underline{\sf From\;1st\;law\;of\;thermodynamics,}\\ \\ \implies \sf \Delta Q = du + \Delta w\\ \\ \\ \sf At\;constant\;volume,\;\Bigg(\dfrac{\Delta Q}{\Delta T}\Bigg)_{v}=\Bigg(\dfrac{du}{\Delta T}\Bigg)_{v}+\Bigg(\dfrac{\Delta W}{\Delta T}\Bigg)_{v}\;\;\;\;\;...........(2)\\ \\ \\ \underline{\sf Now,\;again\;from\;1st\;law\;of\;thermodynamics,}\\ \\ \implies \Delta Q=du+\Delta W$

$\sf At\;constant\;pressure,\;\Bigg(\dfrac{\Delta Q}{\Delta T}\Bigg)_{p}=\Bigg(\dfrac{du}{\Delta T}\Bigg)_{p}+\Bigg(\dfrac{\Delta W}{\Delta T}\Bigg)_{p}\;\;\;\;\;..........(2)$

We know that 'U' do not depands on path or change of method.

$\implies \sf \Bigg(\dfrac{du}{\Delta T}\Bigg)_{p} = \Bigg(\dfrac{du}{\Delta T}\Bigg)_{v}\\ \\ \\ \sf So,\;equation\;(2)\;becomes,\\ \\ \\ \implies \sf C_{p}=\Bigg(\dfrac{du}{\Delta T}\Bigg)_{v} +\Bigg(\dfrac{\Delta W}{\Delta T}\Bigg)_{p}\\ \\ \\ \implies \sf C_{p}=C_{v}+\Bigg(\dfrac{\Delta W}{\Delta T}\Bigg)_{p}\;\;\;\;\;\;...........(3)$

$\sf From\;gas\;equation,\\ \\ \implies \sf Pv=RT\\ \\ \implies P.\Delta V= R.\Delta T\\ \\ \implies \dfrac{P.\Delta V}{\Delta T}=R\\ \\ \implies \dfrac{\Delta W}{\Delta T}=R\;\;\;\;\;............(4)\\ \\ \\ \sf Equation\;(4)\;put\;into\;equation\;(3),\\ \\ \implies {\boxed{\bf C_{p}-C_{v}=R}}$