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Entropy increase in an idealized adiabatic compression

Physical scope

A conditional one-mole monatomic ideal-gas model with positive initial temperature and volume, a sudden constant-load compression to a stated final volume, dissipative settling, and no empirical claim.

Example

This is a conditional model calculation, not a report of an observed compression. The gas is taken to have one mole, $n = 1 m o l$ , of a model whose state variables are a positive temperature $T$ in kelvins, volume $V$ in cubic metres, and uniform pressure $p$ in pascals. The model is monatomic and ideal in the explicit sense that its molar constant-volume heat capacity is $C_{V} = 3 R / 2$ , its internal energy is $U = n C_{V} T$ , and its equation of state is $p V = n R T$ . Its entropy difference is

$S (T, V) - S (T_{0}, V_{0}) = n C_{V} \ln (T / T_{0}) + n R \ln (V / V_{0}),$

for positive reference values $T_{0}$ in K and $V_{0}$ in $m^{3}$ . At initial equilibrium let $T_{i} > 0$ and $V_{i} > 0$ . A piston protocol suddenly raises the external pressure from the initial value $p_{i}$ to a constant final load $p_{f}$ , compresses the gas adiabatically to $V_{f} = (9 / 10) V_{i}$ , and dissipatively settles it to a final equilibrium state. The protocol admits no reverse record, so it is irreversible by Reversible thermodynamic process record. The idealization assumes no change in the included load's entropy; all numerical behavior is a model assumption and deduction, not an empirical measurement.

Facts & Assumptions

Given: The process and sign conventions of Thermodynamic process records and transfer conventions, the closed-record first-law postulate First-law energy balance for closed process records, and the stated ideal-gas model.

[F1]

A process record is adiabatic exactly when $H_{P}^{i n} (t) = H_{P}^{o u t} (t) = 0$ for every $t \in I_{P}$ : no thermal energy crosses its selected outer boundary throughout the record. This implies $Q [P] = 0$ ; $W_{o n} > 0$ denotes work done on the system (Thermodynamic process records and transfer conventions).

[F2]

For every admitted closed record in the stated class, $U (x_{P}^{+}) - U (x_{P}^{-}) = Q [P] + W_{o n} [P]$ (First-law energy balance for closed process records).

[F3]

The adopted entropy is a finite scalar state function with values measured in $J K^{- 1}$ (Concave macroscopic fundamental entropy).

[L1]

$\log$ is strictly increasing, $\log 1 = 0$ , and $\log (x y) = \log x + \log y$ for positive $x, y$ (The natural logarithm as the inverse of the exponential function, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[A1]

The ideal-gas equations and entropy difference are exactly those stated in the Example; $R > 0$ has units $J m o l^{- 1} K^{- 1}$ and $C_{V} = 3 R / 2$ .

[A2]

The gas is the system inside the declared outer boundary: no matter crosses it, its material constitution is fixed, and its mechanical model supplies a total mass $M > 0$ and center-of-mass velocities in the fixed chart of Thermodynamic process records and transfer conventions. The gas center of mass is at rest at both endpoint preparations, so $Δ K_{c m} = 0$ ; no separately tracked external-field potential energy changes (the model may take that term to be absent), so $Δ U_{e x t} = 0$ . Thus the admitted record is in $P_{c l} (s)$ as defined by Thermodynamic process records and transfer conventions. Its protocol has $H_{P}^{i n} (t) = H_{P}^{o u t} (t) = 0$ for every $t \in I_{P}$ , hence $Q [P] = 0$ , a constant external load $p_{f}$ during compression, and $W_{o n} [P] = p_{f} (V_{i} - V_{f})$ . The final equilibrium obeys $p_{f} V_{f} = n R T_{f}$ , and the initial equilibrium obeys $p_{i} V_{i} = n R T_{i}$ . The apparatus is stipulated to dissipatively settle and to have zero entropy change.

Verification

Given: $n = 1 m o l$ , $V_{f} / V_{i} = 9 / 10$ , the premises [F1]–[F3], the logarithm laws [L1], and the model/protocol assumptions [A1]–[A2].

1.1givenF1F2A1A2algebra

Put $λ : = p_{f} / p_{i}$ . The equilibrium equations give $T_{f} / T_{i} = λ (9 / 10)$ . The first-law balance gives $n C_{V} (T_{f} - T_{i}) = p_{f} (V_{i} - V_{f})$ , hence $(3 / 2) (λ (9 / 10) - 1) = λ (1 / 10)$ . Solving yields $λ = 6 / 5$ and $T_{f} / T_{i} = 27 / 25$ .

2.1givenF3L1A1step 1.1algebra∎

The entropy difference is therefore $Δ S = n R [(3 / 2) \ln (27 / 25) + \ln (9 / 10)]$ . By [L1], $2 Δ S / (n R) = \ln ⁣ ((27 / 25)^{3} (9 / 10)^{2}) = \ln (1594323 / 1562500) > 0$ , since $1594323 > 1562500$ and $\log$ is strictly increasing with $\log 1 = 0$ . Since $n R > 0$ , $Δ S > 0$ .

Entropy increase in an idealized adiabatic compression

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources