Thermo 1st law

Question 1

Fundamental thermodynamic relation

Answer 1

dE= đQ + đW
(but: đQ=TdS)
∴dE= TdS + đW

for an ideal gas đW= -PdV so:
dE= TdS - PdV

Question 2

conditions for a process to be reversible

Answer 2

1. must be quasistatic
2. no external friction
3. causes no permanent change (like going over the elastic limit of a spring)

Question 3

1st law of thermodynamics

Answer 3

dE= đQ + đW
The change in internal energy of a system is the sum of the work done on the system and the heat supplied to it.

Question 4

what’s an isothermal/ adiabatic/ isobaric/ isochoric process

Answer 4

isothermal: no temp change (dE= 0 for an ideal gas)

adiabatic: no heat transfer (đQ= 0, dS=0)

isobaric: no pressure change

isochoric: no volume change (đV= 0 so đW=0)

Question 5

relation between đQ and specific heat

Answer 5

đQ= CdT
đQ=dE=CdT (for an ideal gas)

Question 6

prove the relation between Cᵥ and Cₚ

Answer 6

start with the fundamental thermodynamic relation
đQ= dE - đW
CdT= dE + PdV
∴ Cᵥ = ( ∂E/∂T )ᵥ (subscript v means volume is constant so dV=0)

Cₚ = ( ∂E/∂T )ₚ + P( ∂V/∂T )ₚ (subscript p means pressure is constant)
since V= nRT/P, P( ∂V/∂T )ₚ = nR
∴Cₚ = ( ∂E/∂T )ₚ + nR

for an ideal gas E is only a function of temperature so
( ∂E/∂T )ₚ = ( ∂E/∂T )ᵥ
∴Cₚ = Cᵥ + nR

Question 7

Adiabatic index equation

Answer 7

γ= Cₚ / Cᵥ

Question 8

ideal gas relations for an adiabatic process

Answer 8

PV^γ = constant
TV^(γ-1)= constant

note: they are different constants and the constants are different for different processes

Question 9

how do you conceptually understand the way heat is transferred on a PV diagram.

Answer 9

Qₕ enters when the gas is being heated.
Q꜀ exits when the gas is being cooled

Q꜀ also exits when the gas is isothermally compressed (work is done to compress the gas, so it must release heat so that dE=0). and vice versa

Question 10

show that the efficiency of a reversible engine must be less than or equal to the Carnot efficiency

Answer 10

For a reversible engine, the entropy of the cold bath increases and the entropy of the hot bath decreases.
η = w/Qₕ = 1- Q꜀/Qₕ
Q꜀ = T꜀ΔS꜀
Qₕ = TₕΔSₕ

hence: η = 1- T꜀ΔS꜀/TₕΔSₕ

2nd law of thermo: ΔS >= 0
hence: ΔS꜀ >= ΔSₕ

∴ η <= 1- T꜀/Tₕ (which is the Carnot efficiency)

Question 11

Cᵥ for diatomic and monatomic ideal gasses

Answer 11

monatomic:
3/2 * K* T

diatomic:
5/2 * K * T

K: boltzmann constant
T: temp