thermodynamics markscheme

8/13/2019 Thermodynamics Markscheme

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IB Questionbank Physics 1

Thermodynamics Markscheme

1.B

2.B

3.B

4.C

5.B

6.A

7.A

8.B

9.B

10. C

11. B

12. A

13. A

14. A

15. B

16. (a) Q: thermal energy transfer to system;U: change in/difference in internal energy;W: work done by system; 3Accept valid alternative is transferfrom and done on.

(b) (i) isochoric / isovolumetric; 1

(ii) in procedure 1 the gas expands against the atmosphere;this requires extra work to be done;

internal energy change for gas is less;temperature is a measure of internal energy; 4

Ignore references to return of gas to initial state.

(iii)

U / J W / J Q / J

Procedure 1 +120 +80; +200

Procedure 2 +200; 0; +200

3[11]


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17. (a) (i) random motion;no gravitational effect;

no forces of attraction between molecules/atoms;time of collision much less than time between collisions;Newtons laws apply; 2 max

(ii) potential energy not used/irrelevant;

(because) there are no forces between molecules in an ideal gas;gas speeds vary so need to take an average; 3

(b) (i) n= ;

n =0.18mol; 2Award [2]for bald correct answer.

(ii) show use ofpV= constant;

19 105Pa; 2Award [2]for bald correct answer.

(iii) pressure equals ;

(to give 2.8 106Pa)

or

pressure = ;

(c) (i)

smooth curve, curving correct way for (b)(ii);vertical straight line for (b)(iii);smooth curve, steeper than (b)(ii) for (c); 3Labelled curves are not needed as such but direction must be clear.

(ii) (c) identified as H; 1

(iii) recognition that area underneath curve is measure of energy;measure area enclosed by loop /pV changes; 2

[16]

RT

pV

290

1019420 5

4103.2

42031.818.0

V

nRT


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18. (a) (i) ongas is compressed 1 maxCorrect answer and correct explanation.

(ii) ejectedpressure remains constant, volume reducedso temperature must go down 1 max

Correct answer and correct explanation.

(b) work done =pV;

=1.0 105 0.4 =0.40 105J (40 kJ); 2 maxSign should be consistent with (a)(i) above-work by and + work herewould get zero for (a)(i) but [2]marks here.

(c) area enclosed;

0.6 (0.2) 105J (60 kJ 20 kJ); 2 max

(d) efficiency = work out/heat in;

= =50 % (17 %); 2 max

19. (a) isothermal: takes place at constant temperature;adiabatic: no thermal energy exchange between gas and surroundings; 2

(b) (i) neither; 1

(ii) W=PV= 1.2 105 0.05 = 6.0 10

3J; 1

(iii) recognize to use Q= U+ W;

to give U= 2.0 103J; 2

20. (a) (i) no thermal energy into or out of the system / OWTTE; 1

(ii) overall work done BY the gas; 1

(b) B C; any sensible justification; 2e.g. increase in volume at constant pressure requires thermal energy or

temperature increase etc.21. (a) pVconstant for isothermal / adiabatic always steeper;

henceAB; 2

(b) area between lines AB and AC shaded; 1

(c) area is 150 (15) small squares;(allow ecf from (b))

work done = 1.5 1 103

1 105;

= 150 J; 3

For any reasonable approximate area outside the range 150 (

15)squares award [2 max]for the calculation of energy from the area.

(d) no thermal energy enters or leaves / Q= 0;so work done seen as increase in internal energy;hencetemperature rises; 3

Award [0]for a mere quote of the 1st law.

22.a) increase in the degree of disorder (in the system); 1(b) total entropy (of the universe); is increasing; 2

(c) entropy of surroundings increases by a greater factor;because process gives off thermal energy / other appropriate statement; 2

(d) freezing causes release of latent heat;entropy of surroundings increases;

more than the decrease when water freezes; [3]

120

60

thermodynamics markscheme

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