ball, coffee, 32, 33, 39, 43, 49

Ch 21.1      ball

(a) How many atoms of helium gas fill an inflated basketball having a radius of 11.5 cm at 27.0°C and 1.70 atm? (b) What is the average kinetic energy of the helium atoms? (c) What is the root-mean-square speed of the helium atoms?

 (a)     P                  V          = n     R     T (1.013x105)1.7 4π(0.115)3/3 = n 8.314 300 K n = 0.440 moles N = 0.440 moles * 6.022 x 1023 atoms/mole N = 2.65 x 1023 atoms (b)     K = 3/2 kBT           kB = R / NA           K = 3/2 (8.314/6.022 x 1023)(300)           K = 6.22 x 10-21 Joules (c)      ½ mv2 = 3/2 kB  T             mv2    = 3  R/NA T vrms = Ö(3     R     T    / m NA) vrms = Ö(3*8.314*300/0.004 kg) vrms = 1370 m/s   He = 2 atoms with 2 protons & 2 neutrons              by definition: 6.022 x 1023 amu’s = 1.000 g 4amu’s = 0.004 kg

Ch 21.2     coffee

A 2-L Thermos bottle is full of coffee at 95°C.  You pour three cups of coffee and then replace the cap.  Make an order-of-magnitude estimate of the change in temperature of the coffee remaining in the Thermos bottle due to the air that replaces the poured coffee.

 Assume Troom = 20 °C 1 liter ≈ ¼ gallon 1 gallon = 128 ounces 1 liter ≈ 32 ounces One cup is 6 ounces. 1/5 of 1000 ml = 200 ml = 200 cc. Three cups is 600cc. We’ll assume you fill your coffee cup completely so we’ll use 200 cc’s of air (replacing coffee) @ 20 °C and 1400 cc’s of coffee remaining @ 95 °C 1400 cc’s of coffee = 1.4 kg Energy transferred from hot coffee will be transferred to the air that entered Thermos from room. -m      c     ΔT =    +(ρair    Vair)        CV air         ΔTair       mair = ρVair     -1.4 (4186) ΔT = (.0012g/cc*600cc) 1.01 J/(gm K) (≈95 – 20)   ΔT = -0.00931 °C so  ΔT ≈ 10-2 °C N2 = 28 amu;      O2 = 32 amu 80%N2 + 20%O2             = 28.8 amu/molecule 28 amu/molecule = 28.8 g/mole CV air = 7/2 (8.314J/mole K) (mole/28.8g) CV air = 1.01 J/(gram*K)

 Ch 21.3     #32 During the power stroke in a four-stroke automobile engine, the piston is forced down as the mixture of combustion products and air undergoes an adiabatic expansion.  Assume that  (1) the engine is running at 2 500 cycles/min, (2) the gauge pressure right before the expansion is 20.0 atm, (3) the volumes of the mixture right before and after the expansion are 50.0 and 400 cm3, respectively (4) the time involved in the expansion is one-fourth that of the total cycle, and  (5) the mixture behaves like an ideal gas with specific heat ratio 1.40.  Find the average power generated during the expansion. 50 cc = 5x10-5 m3 20 atm (1.013x105 Pa / 1 atm) 20 atm ≈ 20x105 Pa 1 stroke for a 4 stroke @ 2500 cycles/min ¼ (60 sec/1 min)(min / 2500cycles) 0.006 sec per stroke Work = DEint = nCvDT                     Given: Q = 0 J Work = n (5R/2)DT Work = 5/2 (nRDT)                            DPV = nRDT Work = 5/2 (      Pf     Vf      –      P          V ) Work = 5/2 ( 1.14x105 40x10-5 – 20x105 5x10-5 ) Work = -150 Joules   Power = Work / time P = 150 J / 0.006 sec P = 25,000 Watts Assume air plus burnt gasoline behaves like a diatomic ideal gas. PVγ = PfVγf Pf = P  ( V / Vf )γ Pf = 21atm (50/400)1.4 Pf = 1.14 atm

Ch 21.4     #33

Consider 2.00 mol of an ideal diatomic gas. (a) Find the total heat capacity of the gas at constant volume and at constant pressure assuming the molecules rotate but do not vibrate.  (b) What If? Repeat, assuming the molecules both rotate and vibrate.

(a)                                                                         (b)

 N(5)kBT/2   = nCvT nNA(5/2)kBT = nCvT n(5/2)RT = n Cv T n Cv = n (5/2) R n Cv = 2 (5/2) 8.314 n Cv = 41.6 J/K n CP = n (Cv + R) n CP = 2 (5/2 R + R) n CP = 2 (7/2* 8.314) n CP = 58.2 J/K In vibration with the center of mass fixed, both atoms are always moving in opposite directions with equal speeds. Vibration adds two more degrees of freedom for two more terms in the molecular energy, for kinetic and for elastic potential energy. We have n Cv = n (7/2) R n Cv = 58.2 J/K     n CP = n (9/2) R n CP = 74.8 J/K

Ch 21.5     #39

From the Maxwell-Boltzmann speed distribution, show that the most probable speed of a gas molecule is given by Equation 21.29. Note that the most probable speed corresponds to the point at which the slope of the speed distribution curve dNv /dv is zero.

 Maximum values occur when the derivative is set to equal zero (dNV / dV = 0) NV = 4π N(m / 2πkBT)3/2 v2e-E/kBT NV = 4π N(m / 2πkBT)3/2 v2e-mv^2 / 2kBT 4πN (m/2πkBT)3/2    exp(-½mv2/kBT)     (2v – 2mv3/2kBT) = 0 Solve for v to find the most probable speed Reject as solutions  v = 0 and v = ∞ Retain only (2 – mv/kBT) = 0 Then vmp = (2kBT/m)1/2

Ch 21.6     #43

Assume that the Earth’s atmosphere has a uniform temperature of 20°C and uniform composition, with an effective molar mass of 28.9 g/mol.  (a) Show that the number density of molecules depends on height according to nv(γ) = n0 e-mgy / kBT  where n0  is the number density at sea level, where y = 0.  This result is called the law of atmospheres.  (b) Commercial jetliners typically cruise at an altitude of 11.0 km.  Find the ratio of the atmospheric density there to the density at sea level.

 (a) nV(E)  = n0e-  E   /kBT n(h)     = n0e- mgh / kBT n(h)/n0 = e- mgh / kBT (b) n(h)/n0 = e- Mgh / RT M = 0.0289 kg/mole, h = 11,000 m, T = 240 K n(h)/n0 = e-1.600 n(h)/n0 = 0.202 If students have 0.278, ONLY give half credit for this problem, Instructor Solution mistake on temperature. (anything else other than 0.278…like 0.269 or similar…give full credit)

Ch 21.7     #49

Argon gas at atmospheric pressure and 20.0°C is confined in a 1.00 m3 vessel. The effective hard-sphere diameter of the argon atom is 3.10 ´ 10–10 m. (a) Determine the mean free path . (b) Find the pressure when = 1.00 m.  (c) Find the pressure when = 3.10 ´ 10–10 m.

 Equation (from 21.7 notes) l = 1 / √2 πd2nV   P = nVkBT nV = P / kBT   l = 1 / √2 πd2 (P / kBT) l = kBT / √2 πd2 P (a)      l =         kB       T  /  √2 π       d2            P           l = 1.38x10-23 293/ √2π(3.1x10-10)2 (1.013x105)           l = 9.36 x 10-8 meters (b)     P1 l1 = P2 l2                     (All other variables don’t change)           11 11 = P2 (9.36 x 10-8)           P2 = 9.36 x 10-8 atm (c)      P1 l1 = P2 l2           11 (3.1x10-10)1 = P2 (9.36 x 10-8)           P2 = 302 atm

Extra Problems…possibilities for the final exam

A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures?  (c) Find Q, W, and DEint.

 (a)      PVγ = PfVγf           Pf = P  ( V / Vf )γ           Pf = 5atm (12/30)1.4           Pf = 1.39 atm (b)               P = 5 atm (1.013x105 Pa/atm) P Vγ = n R T T =        P               Vγ      /        n        R T = 5.065x105 (12x10-3)1.4 / 2mole(8.314) T = 363 K   PfVγf = n R Tf Tf =    P                  Vγ        / n    R Tf = 1.408x105 (30x10-3)1.4 / 2 (8.314) Tf= 253 K (c)      Given Q = 0 (adiabatic) ΔEint = Q + Work                            so γ = 1.40 and CV = 5R/2 ΔEint = n       CV          ΔT ΔEint = 2 5R/2 (253-365)K ΔEint = -4,660 Joules                     Given:  Q = 0 J so Work = -4,660 Joules

The largest bottle ever made by blowing glass has a volume of about 0.720 m3.  Imagine that this bottle is filled with air that behaves as an ideal diatomic gas.  The bottle is held with its opening at the bottom and rapidly submerged into the ocean.  No air escapes or mixes with the water.  No energy is exchanged with the ocean by heat.  (a) If the final volume of the air is 0.240 m3, by what factor does the internal energy of the air increase? (b) If the bottle is submerged so that the air temperature doubles, how much volume is occupied by air?

 (a)  PVγ = PfVfγ Pf = P (V/Vf)γ Pf = P (.72/.24)1.4 Pf = 4.66P PV/T = PfVf/Tf Tf = Pf         Vf T / P   V Tf = 4.66P(.24)T/P(.72) Tf = 1.55T Internal energy, Eint = nCVT, is directly proportional to temperature, so Eint,f  /  Eint = Tf / T Eint,f  /  Eint = 1.55 (b)     PVγ = PfVfγ Pf/P = (V/Vf)γ PV/T = PfVf/Tf Tf/T = (Pf/P)   (Vf/V) Tf/T = (V/Vf)γ (V/Vf)-1 Tf/T = (V/Vf)γ-1 Tf/T = (V/Vf)γ-1    2  = (.72/Vf)1.4-1 Vf = 0.127 m3

At what temperature would the average speed of helium atoms equal  (a) the escape speed from Earth, 1.12 ´ 104 m/s, and (b) the escape speed from the Moon, 2.37 ´ 103 m/s? (Hint:  Ch13:  escape speed, mHe = 6.64 ´ 10–27 kg.)

 (a) vave2      = 8       kB       T / π   m 112002 = 8(1.38x10-23)T/π6.64 ´ 10–27 T = 23700 K (a) vave2      = 8       kB       T / π   m 23702 = 8(1.38x10-23)T/π6.64 ´ 10–27 T = 1060 K