Heating at constant pressure: the Cp coefficient

Combining the energy balance with the work of the atmosphere yields an important result: the heat capacity at constant pressure $C_p$, and its relation to the one at constant volume $C_v$.

Principle — Specific heat at constant pressure

If an ideal gas is heated while keeping the pressure constant (= atmospheric, thanks to a free piston), the heat required is $\ev{Q = n\,C_p\,(T_B - T_A)}$ with $C_p = C_v + R = \frac{f+2}{2}R$ (Mayer’s relation).

Derivation. At constant pressure the gas law gives $V/T =$ constant, so the expansion is

$V_B - V_A = \frac{nR}{P}(T_B - T_A)$

The work done by the atmosphere (which is compressed) is

$L_\text{atm} = -P\,\Delta V_\text{gas} = -nR(T_B - T_A)$

Substituting into the first law $\Delta E_\text{int} = Q + L_\text{atm}$:

$\frac{f}{2}nR(T_B-T_A) = Q - nR(T_B - T_A)$

Isolating $Q$:

$Q = \left(\frac{f}{2} + 1\right)nR(T_B - T_A) = \frac{f+2}{2}nR(T_B - T_A)$

Summary — Cv and Cp

$\ev{C_v = \frac{f}{2}R \qquad C_p = \frac{f+2}{2}R = C_v + R}$ $C_p > C_v$: at constant pressure more joules are needed because part of the heat goes into work against the atmosphere, not into internal energy. The difference $C_p - C_v = R$ is exactly the molar work done for every kelvin of heating.

Links

Topics: Thermodynamics Concepts: First law of thermodynamics · Thermodynamic transformations · Specific heat and heat capacity · Ideal gas law Skills: Symbolic setup Objects: Ideal gas · Piston and cylinder

Related exercises: Problem — Rectangular cycle in the p-V plane · Same temperature, different heat · Heating under the piston