Air (an ideal gas) is stored in a rigid closed tank of volume V at absolute temperature of T₀ and pressure P. Ignoring the effect of motion and gravity, what will be the specific exergy (availability per unit mass) of air? [Given: Environmental absolute temperature is T0; Environmental pressure is P₀; Gas constant is R.]

Concept:

Exergy is the maximum useful work possible when a system is brought to equilibrium with its surroundings.

For an ideal gas, the specific exergy (availability per unit mass) is given by:

\( e = (h - h_0) - T_0 (s - s_0) \)

For an ideal gas with constant specific heats, the entropy change is:

\( s - s_0 = C_p \ln \frac{T}{T_0} - R \ln \frac{P}{P_0} \)

For a rigid closed tank, volume remains constant, and thus:

\( h - h_0 = 0 \) (since no change in enthalpy in a rigid tank)

Therefore, specific exergy simplifies to:

\( e = - T_0 \left( C_p \ln \frac{T}{T_0} - R \ln \frac{P}{P_0} \right) \)

Using the relation for an ideal gas at constant volume:

\( \frac{T}{T_0} = \frac{P}{P_0} \)

Substituting this into the equation:

\( e = R T_0 \left[ \frac{P_0}{P} - 1 + \ln \frac{P}{P_0} \right] \)