Some useful information for Exam III
1 u = 1.6605 ×10−27 kg 1 cal = 4.186 J 1 kcal = 4186 J = 4.186 kJ
1 atm = 1.013 × 105 N/m2 = 1.013 × 105 Pa = 1.013 bar
= 760 mm-Hg = 760 torr
k = 1.38 × 10−23 J/K
NA = 6.02 × 1023 /mol g = 9.80 m/s2
R = NA k = 8.314 J/(mol·K) = 0.0821 (L·atm)/(mol·K) = 1.99 cal/(mol·K)
T(°C) = (5/9)[T(°F) −32] T(°F) = (9/5) T(°C) + 32 T(K)=T(°C) + 273.15
∆L = α L0 ∆T ∆V = β V0∆T α steel = 1.2 × 10−5 (C°)-1 β steel = 3.6 × 10−5 (C°)-1
α glass = 0.9 × 10−5 (C°)-1 β glass = 2.7 × 10−5 (C°)-1 β water = 21.0 × 10−5 (C°)-1
Y=(F/A)/(∆L/L0) F/A = − Y α ∆T Q = mc∆T = nC∆T c = (1/m) dQ/dT
ciron = 470.0 J/(kg·K) caluminum = 910.0 J/(kg·K) m = nM
Q = ±mL
L ice ¨ water = 79.6 cal/g = 79.6 kcal/kg L water ¨ vapor = 539 cal/g = 539 kcal/kg
cice = 0.51 cal/(g·Co)
cwater = 1.0 cal/(g·Co)
csteam = 0.48 cal/(g·Co)
H = d Q/d t = kA(TH − TC) / L R = L / k H = A eσ T 4 Hnet = A eσ (T 4 – Ts 4 )
σ = 5.67 × 10−8 W/(m2·K4) kglass = 0.8 W/(m·K)
m tot = nM
pV = nRT = NkT
ρ = pM / RT M = NA m T = (273.16 K) lim (p/pt p)
Pt p Ø 0
3kT
3RT
1
3
2
2
=
=
=
(
)
v
v
v
(K )av = m ( )av = kT
r ms
av
m
M
2
2

RT
a
a V

V
=
−
+
− b  = RT
p
p

λ=
2
2 
(V / n) − b (V / n)
(V / n)   n


4π 2r 2 N
3
2
mv
 m  2 2 − 2 kT
f (v ) = 4π 
 v e
 2π kT 
Cv = (3/2) R , (5/2) R , (7/2) R
W=
∫
V2
V1
p dV
v mp =
2kT
m
v av =
8 kT
π m
Cv = 3R
Q = (U2 – U1) + W = ∆U + W
W = p(V2 – V1 )
d Q = n Cp d T
d Q = d U + d W = d U + p d V d Q = n Cv d T
Cv = Mcv
Cp = Cv + R
γ = Cp / Cv = 5/3, 7/5, 9/7, ÿÿÿ pV γ = const.
∆U = nCv (T2 – T1)
W = (Cv / R) (p1V1 – p2V2) = [1/( γ – 1)] (p1V1 – p2V2)
e = W / Q H = 1 − |Q C| / |Q H|
W = Q = Q H + Q C = |Q H| − |Q C|
eOtto = 1 − 1/ r γ −1 r ≡ Vmax / Vmin
|Q C| / |Q H| = TC / TH
K refrigerator or air conditioner = |Q C| / |W |
dS = dQ / T
2
S = ∫ dQ / T
1
eCarnot = 1 − TC / TH
K Carnot = TC /(TH − TC )
∆S = Q / T
S = k ln w