p = RT (11.1) [R=Runiv/mmole]

du = cVdT (11.2)

dh = cpdT (11.3)

cp + cv = R (11.4)

cp = kR/(k-1) (11.6a) [k=]

cv = R/(k-1) (11.6b) [k=]

pvk = p/k = constant (11.12c)

EQUATION OF STATEFOR IDEAL GAS

p = RT (11.1)

Good to 1% for air at 1 atm and temperatures > 140 K (-130 oC) or for room temperature and < 30 atm

= unique constant for each gas[units of Kelvin]

If L doubled (system 2) but same v, then

(# of collisions/sec)1 = v x (1 sec)/L (# of collisions/sec)2 = v x (1 sec)/2L

(# of collisions/sec)1= ½ (# of collisions/sec)2

Daniel Bernoulli ~ Hydrodynamics, 1738

(system 1)

PV = const

Daniel Bernoulli PV = const

p = F/A

F {# collisions / sec}

p1 (# of collisions/sec)1/(L)2

p2 (# of collisions/sec)2/(2L)2

p2 ½ (# of collisions/sec1)/(2L)2

p2 = 1/8 p1

Vol2= 8Vol1

p2Vol2 = p1Vol1 QED

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(# of collisions/sec)1

p1, n1, m1, vx1, T1, L1

(# of collisions/sec)2

p2, n1 m1, vx1, T1, L2=2L1

L 2L

Daniel Bernoulli ~ PV = const

Hydrodynamics, 1738

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“ The elasticity of air is not only increased by compression but by heat supplied to it, and since it is admitted thatheat may be considered as an increasing internal motion of the particles, it follows that … this indicates a more intense motion of the particles of air.”

Daniel Bernoulli

Here was the recipe for quantifying the idea that heat is motion

– two generations before Count Rumford, but it came too early.

IDEAL GAS: p = RT (eq. 1.11)R = Runiv/mmole

pV = N(# of moles)RunivT

Assume perfect elastic reflections so: - 2mvx is change of x-momentum per collision.

Initially assume vx is same for all particles.

What is Pressure ?

t = 2L/vx

= (mvx)/t =2mvx/(2L/vx) = mvx2/L

Time between collisions, t, of particle with samewall is equal to:

L

Force of one particle impact = Magnitude of momentum change per second due to one particle:

nmvx2/L

Magnitude of momentum change per second due to n molecules:

<vx2> = <vy

2> = <vz2>;

<vx2> + <vy

2> + <vz2> = <v2>

<vx2> = 1/3 <v2>

1/3 nm<v2>/L

Pressure = F/A = [1/3nm<v2>/L]/L2

P = 1/3nm<v2>/L3

PV = 1/3nm<v2> = 2/3n (1/2 m<v2>)

average kinetic energy per particle

Empirically it is found that : PV = nkBT

n=#of particles; kB=1.38x10-23 J/K

PV = 2/3n (1/2 m<v2>)

Empirically it is found that:

PV = nkBT

T(Ko) = [2/(3kB) ] [avg K.E.]

pV = (2/3) n <mv2/2>

Uinternal for monotonic gas

Uint = f(T) depending if p or V held constant

uint, v,… designate per unit mass

duint/dT = cv (11.2)

duint/dT =cp

(# of particles)

pV = nkBT

pV = nkBTn = [Nm][NAvag]

6.02x1023

nkBT = Nm x NAvag kBT = Nm x NAvag [Runiv/Navag.] T

pV= NmRunivT

pV= NmRunivT

p=(1/V)Nmmmole{Runiv/mmole}T

p=(m/V){Runiv/mmole}T

p= {Runiv/mmole}T = RT (11.1)

IDEAL GASpV = NmRunivT

p = {Runiv/mmole}Tp v = {Runiv/mmole}TpV Uint = f(T)

The differential work dW done on the gas in compressing it by moving it –dx is –

Fdx.

dW on gas = F(-dx) = -pAdx = -pdVgoes into dT

pV = 2/3 U for monotonic gaspV = (k - 1) U in general

k = cp/cv = 5/3 for monotonic gasU = pV/(k - 1)

dU = (pdV+Vdp)/(k - 1) – eq. of state

ASIDE: Want to derive important relation between p and V for adiabatic condition,

i.e. Q = 0

Compression of gas under adiabatic conditions means all work goes into increasing the

internal energy of the molecules, so:

dU = W = -pdV for adiabatic (Q = 0)

Equation of statedU = (Vdp + pdV) / (k - 1)

Cons. of energy dU = W + Q

dU = W + QdU = (Vdp + pdV) / (k - 1)

-pdV = (Vdp + pdV) / (k - 1)

-(pdV)(k - 1) = Vdp + pdV

-(pdV)k + pdV = Vdp + pdV

-(pdV)k - Vdp = 0

-(pdV)k - Vdp = 0

(divide by -pV)(dV/V) + (dp/p) = 0

(integrate)kln(V) +ln(p) = ln(C)

ln(pVk) = ln(C)

pVk = C or pvk = c (11.12c)

IDEAL GASdU = W + QQ/m = cvdT + pdv pv = RT (R=Runiv/mmole)

pdv + vdp = RdTQ/m = cvdT + RdT - vdpDivide by dT

[Q/m]/dT = cv + R – vdp/dTIf isobaric, i.e. dp=0 then

{[Q/m]/dT}p = cp = cv + R (11.4)

cp = cv + R; cp – cv = RDivide by cv, & let k = cp/cv;

k - 1 = R/cv, or

cv =R/(k-1) (11.6b)

Multiply by cp/cv

cp = kR/(k-1) (11.6a)

usually k = in most books

IDEAL GAS

h = u + pvdh = du + d(pv)dh = cvdT + RdTdh = (cv + R)dTdh = cpdT (11.3)