PhysicsAndMathsTutor.com 1 2 (*) (1)

Size: px

Start display at page:

Download "PhysicsAndMathsTutor.com 1 2 (*) (1)"

Norah Hardy
5 years ago
Views:

1 PhysicsAndMathsTutor.com 1 1. (a) pressure (*) Pa or N m volume m (*) (*) (not allow kpa) number of moles mol (or none) molar gas constant J K 1 mol 1 (mol 1 implies molar) temperature K 4 (b) (i) W(= pv) = = 5.0 J V = constant (or use of pv = nrt) T = 9 T gives T = 440K = = 167 C (iii) any two from the following: speed (or kinetic energy) increases collision rate (with walls) decreases (or time between collisions increases) distance between collisions increases (mean free path inverse prop. to molecular density) 8 [1]. (a) pv = nrt 1 (b) graph to show: straight line through the origin (c) average kinetic energy or speed of the molecules increases with temperature more collisions occur (more frequently) [or more particles hit (per second)] (average) change in momentum (during a collision) is greater rate of change of momentum is greater (hence) force/pressure (during collision) is greater max 4 QWC 1

2 PhysicsAndMathsTutor.com (d) average E k = kt = = J [or use energy per mole = RT = = J mol 1 ] [9]. (a) (i) volume of air is less with the powder present pressure 1/volume so pressure is greater QWC initial volume = (m ) final volume = (m ) final pressure = = Pa P0V 5 0 [alternative: no.of moles (n) ( ) = RT RT nrt 5 0 final pressure = = V = 140 kpa 6 (b) (i) volume of powder = mass 0.1 = m density 700 assuming powder volume as in (b)(i), initial volume = ( ) 10 4 (m ) final volume = ( ) 10 4 (m ) final pressure = = Pa test successful as calculated final pressure = measured final pressure 5 [11] 4. (a) (i) (use of n pv RT gives) n (.1) moles n 9.8 moles

3 PhysicsAndMathsTutor.com (b) (total) ( ) ( ) (allow C.E. for incorrect values of n from (a)) (oxygen molecules) [5] 5. (a) use of pv = constant or p 1 V 1 = p V p = 99.50/4.15 = 8.5 kpa (b) no. of moles = 99 [ 10] / = 1.4() 10 moles no. of molecules (= 1.4() ) = (c) molecules/particles have momentum momentum change at wall momentum change at wall/collision at wall leads to force [allow impulse arguments] less air so fewer molecules so change in momentum per second/rate of change is less [or per unit per time] pressure is proportional to number of molecules (per unit volume) max 5 [11] 6. (a) energy produced per minute = (J) (= ) 5 volume of gas required =.7 10 (m ) (= m ) [or volume of gas per sec 4500 = (m ) volume of gas per minute = (m ) (= m )] (b) (i) ( m gives) m (g) = V = kg mass of one molecule = A (= kg) N

4 PhysicsAndMathsTutor.com 4 number of molecules = = (allow C.E. for value of m g ) [or number of moles 5 10 = number of molecules (= ) = (allow C.E. for number of moles) 4 (c) energy per minute (from (a)) = (J) Q 5 (use of Q = mc θ gives) m (= ) =.7 10 cθ 990 (614) = 1(.4) kg [or 4500 m for 1 second = 0.1 (kg for 1 second) 990(614) (= ) = 1 kg for 1 minute ] (d) (use of P = IV gives) I P = = 19(.6) A V 0 current required for 4.5 kw would exceed fuse rating [or P = kw for 1 A] [11] 7. (a) (i) thermal energy gained by water = (5 15) = J (thermal energy loss by copper = thermal energy gained by water gives) ΔT = ΔT = = 808 K flame temperature (= ºC) = 84ºC or 1116 K

5 PhysicsAndMathsTutor.com 5 (b) (i) measure the total mass of the water, beaker and iron lump (to find the mass of water lost) mass of water lost due to conversion to steam, m = mass measured in (b) (i) - initial mass of water, beaker and iron add the thermal energy due to steam produced, ml, to the thermal energy gained by the water calculated flame temperature would be greater 4 [7] 8. (a) (i) (use of pv = nrt gives) =0.194 moles n average kinetic energy = / = J 4 (b) same temperature (hence) same average kinetic energy (since) kinetic energy is 1 mv nitrogen molecules must have a higher mean square speed [7] 9. (a) (i) pv = nrt all particles identical or have same mass collisions of gas molecules are elastic inter molecular forces are negligible (except during collisions) volume of molecules is negligible (compared to volume of container) time of collisions is negligible motion of molecules is random large number of molecules present (therefore statistical analysis applies) monamatic gas Newtonian mechanics applies max 4

6 PhysicsAndMathsTutor.com 6 (b) E k = RT or = N A kt = J ( J) (c) masses are different hence because E k is the same, mean square speeds must be different [9] 10. (a) (i) no net flow of (thermal) energy (between two or more bodies) bodies at same temperature (kinetic) energy is exchanged in molecular collisions until average kinetic energy of all molecules is the same max RT (b) (i) c r.m.s. M = 140m s 1 = average k.e. of nitrogen molecules = average k.e. of helium molecules = (140) = J alternative schemes for : average k.e. = kt = = J

7 PhysicsAndMathsTutor.com 7 or RT average k.e. = N A = = J nkt (iii) use of p = or equivalent [or, at same temperature, V p no. of molecules] p He = 10 = 80 kpa 6 [9] 11. (a) curve A below original, curve B above original both curves correct shape (b) (i) (use of pv = nrt gives) = n n = 11 (mol) (10.8 mol) (use of E k = kt gives) E k = = J (iii) (no. of molecules) N = (= ) total k.e. = = J (allow C.E. for value of n and E k from (i) and ) (use of n = 11 (mol) gives total k.e. =.9 (7) 10 4 J) 5 [7]

8 PhysicsAndMathsTutor.com 8 1. (a) (i) a collision in which kinetic energy is conserved molecules of a gas are identical [or all molecules have the same mass] molecules exert no forces on each other except during impact motion of molecules is random [or molecules move in random directions] volume of molecules is negligible (compared to volume of container) [or very small compared to volume of container or point particles] time of collision is negligible (compared to time between collisions) Newton s laws apply large number of particles (any two) (b) (i) the hot gas cools and cooler gas heats up until they are at same temperature hydrogen molecules transfer energy to oxygen molecules until average k.e. is the same (any two ) (use of E k = kt gives) Ek = = J ( J) 4 [7] 1. (a) p 1 V 1 = p V (0.0 + V) = V V = 41.5 cm (b) rapid compression gives little time for heat to leave canister adiabatic process air temperature rises heat is lost temperature (and pressure) fall [Max 5]

9 PhysicsAndMathsTutor.com (a) (i) T (=7 + ) = 95 (K) pv = nrt = n n = 1160 (moles) (1156 moles) (allow C.E. for T (in K) from (i) (iii) N = = ( ) 5 (b) (i) decreases because temperature depends on mean square speed (or [or depends on mean E k ] decreases as number of collisions (per second) falls rate of change of momentum decreases [or if using pv = nrt decreases as V constant as n constant ] [or if using p = 1/ρ c decrease as ρ is constant as c ) c is constant ] max 4 [9] 15. (a) p: pressure and V: volume N: number of molecules m: mass of one molecule/particle/atom c : mean square speed 4 (b) (i) molecules have a range of speeds they have no preferred direction of movement elastic collisions intermolecular forces are negligible (except during collisions) volume of molecules negligible (compared to volume of container) time of collisions negligible (compared to time between collisions) all molecules identical laws of statistics apply or large number of molecules Newtonian laws apply any two max

10 PhysicsAndMathsTutor.com 10 (c) molecules collide (with the walls) walls exert a force on the molecules molecules exert an (equal) force (on the walls) creating pressure molecule momentum changes max 4 [11] 16. (a) (i) p V = nr T V = (gives V = m ) (use of E k = kt gives) Ek = = (J) 4 (b) (use of pv = nrt gives) n = [or use p n] n = 1 moles (1.5 moles) (c) pressure is due to molecular bombardment [or moving molecules] when gas is removed there are fewer molecules in the cylinder [or density decreases] (rate of) bombardment decreases molecules exert forces on wall c is constant 1 [or pv = Nm c V and m constant c constant since T constant p N ] 1 [or p = c explanation of decreasing ] c constant since T constant p ] max 4 [10] 17. (a) (i) more collisions (with wall) per second and more momentum change per collision greater force because more momentum change per

11 PhysicsAndMathsTutor.com 11 second (greater pressure) (same pressure) faster molecules so more momentum change per collision greater volume, fewer collisions per second [or greater volume to maintain same pressure] max 4 nrt (b) (i) V = p = = m T = T1V p p1v1 = pv or T nr = 1000 K (77 C) 4 [8] 18. (a) (i) correct p and V from graph 4 n = (= mol) T V = V 1 T1 =. 10 m (b) (i) RT or NA kt total kinetic energy nrt = = 9 J molecules have no potential energy no attractive forces [or elastic collisions occur] max 4 (c) Q = heat entering (or leaving) gas U = change (or increase) in internal energy W = work done [ for three definitions, deduct one for each incorrect or missing]

12 PhysicsAndMathsTutor.com 1 (i) Q = U temperature rises but no work done Q = U +W temperature rises and work done in expanding max 5 (d) (i) U = nr(500 00) = 159 J (= Q) pv = (..0) 10 = 104 J Q = U+ pv = 6 J [15] 19. (a) (i) pressure (average) kinetic energy [or rms speed] (b) (i) pv = nrt n = = 1.0(mol) (1.4 mol) mass of air = = 0.06 kg (allow e.c.f from(i)) (iii) = 1 10 =.6 kg m (allow e.c.f. from) 5 (c) (i) same because the temperature is the same QWC different because the mass of the molecules are different 4 [11]

PhysicsAndMathsTutor.com 1

PhysicsAndMathsTutor.com 1 Q1. In an experiment to measure the temperature of the flame of a Bunsen burner, a lump of copper of mass 0.12 kg is heated in the flame for several minutes. The copper is then