Q.(a) Derive Mayer's relation for an ideal gas. OR

Applying the first law of thermodynamics to one mole of ideal gas at constant volume and at constant pressure gives Mayer's relation: Cp - Cv = R.

This question offers an OR alternative (horizontal oscillations of a spring); this solution answers the primary part (a) as instructed.

DERIVATION OF MAYER'S RELATION:

Consider one mole of an ideal gas. By the first law of thermodynamics, the heat supplied dQ to a system equals the increase in its internal energy dU plus the work done by the gas dW:

dQ = dU + dW, where dW = P dV

Case 1 -- Heating at CONSTANT VOLUME:

At constant volume, dV = 0, so no work is done (dW = 0). All the heat supplied goes entirely into increasing the internal energy:

dQ_v = dU

By definition, the molar specific heat at constant volume, Cv, is the heat required to raise the temperature of one mole by dT at constant volume:

dQ_v = Cv dT

So: dU = Cv dT ... (1)

(Since the internal energy of an ideal gas depends only on temperature, this relation dU = Cv dT actually holds for ANY process the ideal gas undergoes, not just a constant-volume one.)

Case 2 -- Heating at CONSTANT PRESSURE:

At constant pressure, the gas does work as it expands: dW = P dV. The heat supplied at constant pressure, by definition of Cp, is:

dQ_p = Cp dT

By the first law:

dQ_p = dU + P dV

Cp dT = Cv dT + P dV ... using (1) for dU

For one mole of an ideal gas, the equation of state is:

PV = RT

At constant pressure, differentiating: P dV = R dT

Substituting:

Cp dT = Cv dT + R dT

Dividing throughout by dT: …