PHYSICS 218 SOLUTION TO HW 8. Created: November 20, :15 pm Last updated: November 21, 2004

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1 Created: November 20, :5 pm Last updated: November 2, Schroeder.6 (a) The three forces actin on the slab of thickness dz are ravity and the pressure from above and below. To achieve equilibrium they must add to zero. 0= mρadz [P (z + dz) P (z)]a () Expandin P(z) in a Taylor series yields P (z + dz) =P (z)+ P/ z dz = P (z)+ (z) and after division by Adz we thus find: = mρ (2) dz (b) The ideal as law is PV = Nk B T ρ =(mp )/(k B T ) insertin ives the barometric equation dz = m k B T P (3) (c) For a heiht independent temperature we use an exponential ansatz P (z) =P (0)e λz that yields dz = λp (0)eλz = m k B T P (0)eλz (4) This aain ives λ = (m)/(k B T ) and thus P (z) =P (0)e mz/kt (5) Substitutin this function for P (z) into the formula for ρ = mp/kt ives us ρ(z) = mp (0) kt e mz/kt = ρ(0)e mz/kt (6) (d) The numerical value for m/kt is equal to /m for an averae molecule mass 28.8u and a temperature of 293K. Thus we find Oden, Utah Leadville, Colorado Mt. Whitney, California Mt. Everest, Nepal/Tibet P / atm However we have to bear in mind that these values are approximate. Especially for the Mt. Everest we would find a very different value if we took the temperature decrease into account. 2. Schroeder.22 (a) Each molecule hittin the Area A will on averae exert a force of 2mv x, dividin the total force by this amount we find an approximation for the averae number of molecules hittin the wall. N = PA 2mv x t (7) (b) Usin ( /2 vx) 2 as an approximation to vx we use v 2 = vx 2 + vy 2 + vz 2 = 3vx 2 = 3kT/m to find v x kt/m.

2 (c) Usin the ideal as law to substitute P and (b) we find dn dt = PA 2mv x = A kt 2V m N (8) Aain an exponential ansatz N(t) =N(0)e t/τ will lead us to the result τ A kt N(t) = 2V m N(t) τ = 2V m A kt (9) (d) Assumin the same data for air as in () we find the characteristic time τ =6.88s. (e) Assumin a diameter of 28 and a radius of one inch we find a volume of m 3. Estimatin the pressure in the tire at the beinnin of the leakin to be about 4atm we can infer from the ideal as law that the density inside the tire is four times hiher than outside at sea level. We thus need that the number of molecules drops to a fourth. t τ = lo 4.38 (0) Now solvin for the size of the hole we have A = 2V τ m kt 2V.38 m = t kt = mm 2 () (f) We make the assumptions that the window with the do in it leaves about 000cm 2 uncovered, that the capsule miht approximately have a volume of 20m 3 and that it takes our adventurers about one second to et rid of the do. Assumin air at room temperature in their capsule we find τ =3.75s N final e /3.7 =0.39 (2) N initial We miht thus conclude that despite our rather positive assumptions this inenious method is not a realistic way to et rid of the do corpse. However it is not as bad as one miht suspect. 3. Schroeder.3 3 V / l final initial (a) 3 P / atm (b) With the mentioned assumption P (V )=c V and P i V i = 0.3J we find Vf 3Vi W = P (V )dv = cv dv = c V i V i 2 [9V i 2 Vi 2 ]= 4P i V i = 405.2J (3) The dimensions for the constant need to be [c] =N/m 5. 2

3 (c) Knowin the ideal as law we have PV = NkT =(2/f)U. In the case of Helium, a monatomic as, f =3. Wehave U i = 3 2 P iv i U f = 3 2 9P iv i =9U i (4) and thus U =8U i =2P i V i = 25.6J. (d) We know from enery conservation U = Q + W and thus Q = U W = (2 ( 4))P i V i = 620.8J (5) (e) To realize a pressure risin proportional to the volume we need some device whose force increases linearly. A possible example is a cylinder with movable piston, if we connect the piston to a sprin. The volume increase leads to a linear decrease in the lenth of the sprin. This decrease via Hooke s law to a linearly risin pressure, if all other walls of the cylinder are fixed. 4. Schroeder.32 P/atm 200 V / l 0.99 The work needed is Vf W = V i P (V )dv, (6) the area under the curve in the PV-diaram. In our case however the volume occupied by the water almost does not chane at all. If we assume that the pressure rises linearly with the volume reduction the area under the curve is in ood approximation a trianle. W = m N/m 2 J (7) This is indeed very small, but we need to remember that work is force multiplied by distance. Here the latter one is obviously very small. 5. Schroeder.34 (a) We can always use U = Q + W = f/2 Nk T = f/2 V P, where the latter equality follows at fixed volume from the ideal as law. We find the derees of freedom to be = 5, where we only have two rotational derees of freedom as the rotation about the axis of the molecule is not excited for quantum mechanical reasons. A There is no chane in the volume, thus U =2.5V (P 2 P )=Q (8) 3

4 B The pressure is fixed, thus as T = p/nk V and and finally Q = U W =3.5P 2 (V 2 V ). C Aain no chane in volume, thus D Aain the pressure is fixed, thus U = f/2 V V =2.5P 2 (V 2 V ) (9) W = P 2 V = P 2 (V 2 V ) (20) U = 2.5V 2 (P 2 P )=Q (2) U = 2.5P (V 2 V ) and W = P (V 2 V ) (22) and Q = U W = 3.5P (V 2 V ). (b) Durin step A we fix the piston and heat the as. In step B we keep the force on the piston constant, leadin to an expansion of the as while we still add some heat. In C we cool the as until the force on the fixed piston is aain P. Finally in step D we compress the as, absorbin the amount of heat necessary to keep the pressure fixed. (c) U i = 2.5(V 2 V )(P 2 P )+2.5(P 2 P )(V 2 V ) = 0 (23) A,B,C,D This is expected, as the internal enery for an ideal as is only determined by its temperature. But the temperature does not chane under one whole cycle, as this means that the system returns to the initial values of the state variables P and V. A,B,C,D W i =( P 2 + P )(V 2 V ) (24) and from 0 = U = Q + W it follows Q =(P 2 P )(V 2 V ). We find that we have added heat to the system and the system in turn has performed work on the piston. 6. Schroeder.36 (a) Somce, by assumption, the process is adiabatic, the quantity PV γ is conserved. For air we have γ =(5+2)/5 =.4. P V γ =7P V γ 2 V 2 =0.249V =0.249l (25) (b) As no heat is exchaned, the work is equal to the chane in internal enery. W = 5 2 Nk(T 2 T )= 5 2 (P 2V 2 P V )= J = 88J (26) 2 (c) From the ideal as law we have that P γ T γ is another conserved quantity. Thus P 0.4 T.4 =(7P ) 0.4 T.4 2 T 2 =7 2/7 T =.74T = 522K (27) 4

5 7. Schroeder.4 (a) To heat 250 of water by 4K we use Q = C p T and assume that C p is rouhly constant and equal to 4.87J/K in this reime. Thus for our 250 we have C p = 047J/K. The amount of heat ained by the water is thus Q H2 O = 488J. (b) As we assume that there is no heat exchane with the styrofoam cup or the atmosphere we know immediately that Q metal = 488J. (c) We may assume that after a minute the metal is in thermal equilibrium with the water. Thus T =76K. Usin the same formula as in (a) we determine C p (metal) = 488J/76K =55J/K. (d) The metal used in this experiment has a specific heat capacity of 0.55J/K. 8. Schroeder.47 We add an amount of ice to absorb heat from the boilin tea so that ice and tea eventually reach thermal equilibrium at 65 deree Celsius. In this process the heat that can be absorbed by one ram of ice is Q = 7.5cal +80cal +65cal = 52.5 cal where the first contribution is the heat capacity of ice, the second one the latent heat and the last one the heat capacity of water. It is interestin to note that it takes almost as much heat to melt the ice at it takes to brin water to boilin. To cool our tea we need Q = 7000cal. Since we assume that the system does not exchane heat with its surroundins, the total chane in internal enery is zero. Obviously there is no work done either, thus (28) 0 = U = Q = x52.5cal/ +( 7000cal) x = 45.9 (29) 5