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Maud Gibson
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1 ABCD BEF F F D CC

10 Vetri Velan GSI, Physics 7B Midterm 1: Problem 3 3. a) Since the collisions with the walls are all elastic, for each given wall, only one component of the velocity matters. Consider the left wall of the box. Collisions with it will only change the x-component of the velocity, while the y-component will be invariant. Thus: t = x = 2l v x v x This is the same for all walls so: t = 2l v x = 2l v y Rubric: 2 pts for using t = x v 2 pts for using the correct velocity (i.e. only v x or v y ) 2 pts for using the correct x = 2l 2 pts for correct answer b) Per particle, F = dp dt p t = 2mv x t For the entire system, F = mv2 x l F tot = N mv2 x l = mv2 y l = N mv2 y l Rubric: 2 pts for realizing F = dp/dt 1 pt for approximating this by F = p t 2 pts for using correct p and t (if they got previous question wrong but used same t that is fine) 1 pt for multiplying by N c) There is an issue with this question. The temperature definition is given by T = mv2 k B but this is per degree of freedom of velocity. So the equation is better written as T = mv2 x k B. However, based on the absolute velocity (which is defined in Part A), it is actually T = mv2 2k B. Thus students answers will be off by a factor of 2 depending on which definition they use. I will proceed with the 1

11 solution using the correct equation. T = mv2 x k B = F totl Nk B Nk B T = Fl = F l l2 P A = Nk B T where P is like an effective pressure, except that it is F/l instead of F/A, and A is the area of the box. If the students had used the wrong equipartition theorem (as given in the exam), they would get: 2P A = Nk B T Rubric: 3 pts for converting T equation into equation of F,l, etc. (if they got previous question wrong but used the correct math and logic here, that is fine) 1 pt for correct v = v x or = 2v x 1 pt for realizing it is 2D (not using volume or pressure, because those are not the appropriate quantities) 1 pt for correct answer No points deducted for getting equipartition wrong by a factor of 2 (see above) 2

12 Problem 4 (a) V 1 = V 0 (1+β T) so that (r 1 ) 3 = (r 0 ) 3 (1+β T). Expanding to first order, we have +2 points for volume expansion equation +2 points for radius cubed formula or +5 points for just right answer r 1 = r 0 (1+β T/3) (1) (b) We assume that the gas is compressed adiabatically so that PV γ 0 = P 1V γ 1, where γ = 5/3, and we obtain ( ) γ ( V0 R 3 r 3 ) γ 0 P 1 = P = P V 1 R 3 r1 3 (2) +2 points for adiabatic relationship +2 points for correct volumes +1 for correct answer (c) Using the above relation between pressure and volume, one finds that T 0 V γ 1 0 = T 1 V γ 1 1 so that +3 points for adiabatic relationship +2 points for using the same volumes as part (b) ( ) γ 1 ( V0 R 3 r 3 ) γ 1 0 T 1 = T 0 = T 0 V 1 R 3 r1 3 (3) (d) Assuming that the particles move in three dimensions, we have v rms = 3kT1 m (4) If the particles move radially, they only have one degree of freedom, so m v r = kt and we have v r = kt1 m (5) + 3 for correct expression of v rms +2 for correct expression of v r 1

Problem 5-Solutions (a) Find P2 and V2 (3points total) During process A, the volume of the gas is held constant therefore: V 2 = V 1 (2points) Also, the problem states that point 2 is at four time

13 Problem 5-Solutions (a) Find P2 and V2 (3points total) During process A, the volume of the gas is held constant therefore: V 2 = V 1 (2points) Also, the problem states that point 2 is at four time the initial pressure of the gas so: P 2 = 4P 1 (1points) Find P3 and V3 (7 points total) From the adiabatic process B we know that substituting for P2 and V2 as we found above: From the isothermal process C we know that P 3 V γ 3 = P 2V γ 2 P 3 V γ 3 = 4P 1V1 γ eqn1 P 3 V 3 = P 1 V 1 eqn2 therefore from eqn1 and eqn2 we can conclude: V γ 1 3 = 4V1 γ 1, V 3 = 4 1/γ 1 V 1 where γ = d+2 d = 7 5 (1point) for a diatomic gas so V 3 = 32V 1 (3points) Finally, P3 can be found from eqn2 to be: P 3 = P 1V 1 32V 1 = P 1 32 (3points) (b) We know that process A is isovolumetric (W A = 0), process B is adiabatic (Q B = 0)and process C is isothermal ( E C = 0): 1

14 E Q W Process A E A Q A 0 Process B E B 0 W B Process C 0 Q C W C Using the first law of thermodynamics ( E = Q W) for process A: Q A = E A Also, we know that E t otal for the cycle should be 0, therefore E B = E A Also, using the first law of thermodynamics for process B: W B = E B = E A So all we need is E A to solve for Q A, E B, W B : E A = d 2 nr(t 2 T 1 ) from the ideal gas law we can find the relation between T 1 and T 2 : P 1 V 1 T 1 = P 2V 2 T 2 = 4P 1V1 T 2 T 2 = 4T 1 E A = d 2 nr(3t 1) where d = 5 2 for a diatomic gas so E A = 15 2 nrt 1, Q A = 15 2 nrt 1, E B = 15 2 nrt 1, W B = 15 2 nrt 1 For process C, we don t know Q C but we can calculate W C W C = V3 =32V 1 V 2 =V 1 PdV Since P is not constant, we can write it in terms of T using the ideal gas law: W C = nrt 1 V2 =V 1 V 3 =32V 1 1 V dv = 1RT 1ln32 Finally, using the first law of thermodynamics for process C: W C = Q C = RT 1 ln32 E Q W 15 Process A 2 nrt nrt 1 0 Process B 15 2 nrt nrt 1 Process C 0 RT 1 ln32 RT 1 ln32 1 point for each E,W and Q (9 total), and 1 point for d =

Lecture 7: Kinetic Theory of Gases, Part 2. ! = mn v x

Lecture 7: Kinetic Theory of Gases, Part 2 Last lecture, we began to explore the behavior of an ideal gas in terms of the molecules in it We found that the pressure of the gas was: P = N 2 mv x,i! = mn