In a rigid insulated box

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · The laws of thermodynamics · The laws of thermodynamics

Thermal Physics, Ch The second law, entropy and irreversibility

Problem

In a rigid insulated box,

n = 2.00\ \mathrm{mol}

of argon, treated as an ideal monatomic gas with constant

C_V = \tfrac32 nR

, is confined to

V_1 = 15.0\ \mathrm{L}

T_1 = 320\ \mathrm{K}

; the rest of the box, of volume

15.0\ \mathrm{L}

, is empty. The partition is removed and the gas fills

V_2 = 30.0\ \mathrm{L}

. Since the box is rigid and insulated and the gas pushes on nothing,

Q = 0

and

W = 0

for this real process, and the final temperature is again

320\ \mathrm{K}

. Take

R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}

. ((a)) Starting from the first law

\mathrm{d}U = \delta Q + \delta W

with the quasistatic work

\delta W = -p\,\mathrm{d}V

and the definition

\mathrm{d}S = \delta Q_{\rm rev}/T

, derive the thermodynamic identity

\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V

, and say in one or two sentences why the result relates any two neighbouring equilibrium states, whatever process actually joined them. ((b)) Find

\Delta U

for the free expansion above. ((c)) Find

\Delta S

of the gas. ((d)) Along a reversible isothermal path at

320\ \mathrm{K}

joining the same two end states, evaluate

\int T\,\mathrm{d}S

and

\int (-p)\,\mathrm{d}V

, and check that their sum is the

\Delta U

of part (b). Answer in the order:

\int T\,\mathrm{d}S

, then

\int(-p)\,\mathrm{d}V