One mole of helium

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

One mole of helium, treated as an ideal monatomic gas with

C_V = \tfrac32 R

constant, is taken round a Carnot cycle between a hot reservoir at

T_{\rm h} = 600\,\mathrm{K}

and a cold reservoir at

T_{\rm c} = 300\,\mathrm{K}

. Both reservoirs are large enough that their temperatures do not change. The four legs are quasistatic: isothermal expansion at

600\,\mathrm{K}

from

V_1 = 5.00\,\mathrm{L}

V_2 = 15.0\,\mathrm{L}

, adiabatic expansion to

300\,\mathrm{K}

, isothermal compression at

300\,\mathrm{K}

, and adiabatic compression back to state 1. Heats are counted as delivered to the gas, so a rejected heat is negative. ((a)) Find the heat delivered to the gas on the hot isotherm and on the cold isotherm (use the adiabatic relation

TV^{\gamma-1} = \text{constant}

to get the volumes on the cold isotherm), and evaluate

\oint \delta Q/T

round the cycle. ((b)) A real engine is now run between the same two reservoirs. Per cycle it takes the same heat from the

600\,\mathrm{K}

reservoir as in part (a) but delivers only

2100\,\mathrm{J}

of work, its working substance returning to the same state each cycle. Evaluate

\oint \delta Q/T

for this cycle, with each heat divided by the temperature of the reservoir it came from, and find the entropy created per cycle. ((c)) The coldest available body is the

300\,\mathrm{K}

reservoir. Find the work lost per cycle, and check it against the work the reversible engine of part (a) delivers from the same hot heat.

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