Heating with a movable piston (and the atmosphere)

If the container has a movable piston that keeps the pressure equal to atmospheric pressure, the gas expands as it heats up. Now three contributions appear in the balance: the change in internal energy, the heat supplied and the work done by the atmosphere.

As the gas expands by $\Delta V_\text{gas} > 0$, the atmosphere is compressed by the same amount, i.e. $\Delta V_\text{atm} = -\Delta V_\text{gas} < 0$. The work the atmosphere does on the gas is therefore

$L_\text{atm} = P_\text{atm}\,\Delta V_\text{atm} = -P_\text{atm}\,\Delta V_\text{gas} < 0$

The negative sign should be read in terms of the gas’s balance: the atmosphere gives energy to the gas when it compresses it, but here it is the gas that pushes the atmosphere out, so this term subtracts available energy. Since there is nothing above the piston but the atmosphere, the gas does no work on other objects: $L_\text{gas}=0$. The first law becomes

$\Delta E_\text{int} = Q + L_\text{atm}$

More heat is needed than at constant volume

For the same temperature change, with a movable piston more heat is needed: part goes into increasing the internal energy, the rest into “pushing away” the atmosphere. This is the physical origin of the fact that $C_p > C_v$.

The complete numerical calculation (nitrogen under a free piston) is worked through in Heating under the piston; the general result $C_p = C_v + R$ is derived in Heating at constant pressure: the coefficient Cp.

Connections

Topics: Termodinamica Concepts: Primo principio della termodinamica · Energia interna · Trasformazioni termodinamiche Objects: Gas ideale · Pistone e cilindro

Related exercises: Worked example — reversible vs irreversible isothermal expansion · Nitrogen heated in a rigid cylinder · Same temperature, different heat