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Chapter 02 · Part III · Thermodynamics
Revised on ____________________
Chapter 02 · The Laws of Thermodynamics

Heat in, work out, and the price of the exchange

2 sheets
2 diagrams
28 results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

I

The first law

W IS WORK DONE BY THE GAS
I.First law  ΔU = Q − W state the convention and never switch
II.Work  W = ∫ P dV area under the P–V path
III.Internal energy  ΔU = n CV ΔT any process, ideal gas
IV.Sign of W  expansion positive, compression negative
V.Isobaric  W = P ΔV = n R ΔT
VI.Isothermal  W = n R T ln( V2V1 ) ΔU = 0, so Q = W
VII.Adiabatic  W = P1V1 − P2V2γ − 1 Q = 0, so ΔU = −W
VIII.Isochoric  W = 0 Q = ΔU = n CV ΔT
PVstartisobaricisochoricisothermaladiabatic
II

Ideal gas and specific heats

γ = 1 + 2/f, AND NOTHING ELSE
I.Equation of state  P V = n R T
II.Mayer's relation  CP − CV = R
III.Ratio  γ = CPCV = 1 + 2f
IV.Adiabatic relations  P Vγ = const , T Vγ−1 = const
V.Monatomic / diatomic  γ = 5/3 , γ = 7/5
VI.Slope on P–V  adiabatic is γ times steeper than isothermal
Q = n CV ΔT holds at constant volume only. ΔU = n CV ΔT holds always — that one line separates the students who finish this chapter from the ones who do not.
III

Cycles and engines

ROUND A LOOP, ΔU = 0
I.Over a cycle  ΔU = 0 so Qnet = Wnet
II.Work done  W = area enclosed clockwise: engine; anticlockwise: fridge
III.Efficiency  η = WQH = 1 − QCQH
IV.Carnot limit  η = 1 − TCTH kelvin, and no engine beats it
V.Refrigerator  COP = QCW = TCTH − TC
VI.Heat pump  COP = QHW = COPfridge + 1
VII.Carnot cycle  two isotherms, two adiabatics reversible throughout
VIII.Free expansion  W = 0, Q = 0, ΔT = 0 no load to push against
PVTHTCABCD2 isotherms2 adiabats
IV

The second law and entropy

THE DIRECTION OF TIME
I.Kelvin statement  no cycle converts heat wholly into work
II.Clausius statement  heat does not flow cold → hot unaided
III.Entropy  ΔS = ∫ dQrevT along any reversible path between the states
IV.Isothermal change  ΔS = n R ln( V2V1 )
V.Heating at constant V  ΔS = n CV ln( T2T1 )
VI.Universe  ΔS ≥ 0 equality only if reversible
V

Where the marks go

FOUR ERRORS, EVERY YEAR
Where marks are lost in this chapter
1Importing the chemistry convention ΔU = Q + W. Here W is work done by the gas: ΔU = Q − W.
2Writing Q = n CV ΔT for an adiabatic compression. Q = 0 there, whatever ΔT is.
3Reading “isothermal” as “no heat flows”. ΔU = 0 gives Q = W; it is adiabatic that gives Q = 0.
4Celsius in η = 1 − TC/TH. The round-number answer is the lure — convert to kelvin first.