Answers to Properties of Matter Homework

  Density 

 1.	r = 7.7g/cm3 or 7,700kg/m3		
 2. 	r = 9,142.86kg/m3			
 3. 	V = 0.02378m3	
 4.	m =1.36x10-3kg    (NB:100ml = 100cm3 = 100x10-6m3)	
 5.	V = 1.0017x10-3m3 = 1001.7ml
 6.	83.6mm
 7.	Lithium would be 42.36 times longer than osmium.

 8.	The particle spacing in solids and liquids of the same
	substance is about the same. However, in the gaseous state
	the particle spacing is 10x greater. This means that for a given
	mass of a substance the solid and liquid would occupy roughly the 
	same volume and have an equal density, but the same mass of gas 
	would occupy a volume 1000x greater (V=10x10x10) and have a density 
	1000x less. (r = m/V)

 9. 	E		(1993 PI  Q11)
 10.	1.103kg/m3	(1993 PI  Q32)

 11.	11.9
 12.	Low density polystyrene displaces higher density water from the ship.
	Reducing the overall density of the ship to less than that of the water 
	will make the ship float.

 13.	Weight of 1m3 of lead = mg =107800N
	To make the lead float in mercury this means that the
	mercury must produce an upthrust of 107800N.
	

	The upward pressure on the lower surface of the lead,
	produced by the mercury liquid, must produce an upward 
	force that will balance the weight.

	Fliquid = PliquidA = (rgh)A
	Fliquid = weight
	(rgh)A = mg
      =>h = m/rA
      =>h = (11,000kg/m3)/[(13600kg/m3)x(1m2)]
      =>h = 0.8088m
 	
  Pressure 

14.	P = 2,450Pa
15.	F = 6237N
16.	E		(1992 PI  Q10)
17.	E		(1995 PI  Q9)
18.	B		(1999 PI  Q8)
19.	C		(1996 PI  Q10)
20.	m = 50kg
21.	F =176,000N
22.	A = 1x10-2m2 (using g=10N/kg)
23.a.	A sharp blade has a small surface area.
	This creates a large pressure when a force
	is applied to the knife because: P = F/A.

   b.	Skis increase the contact area with the snow. 
	The pressure produced by the weight of the person
	is therefore reduced (P = F/A).

24.	P = 18849.56N
25.	Fres = 15x104N  (to the left)		(1993 PI  Q32)

Pressure in a Fluid	 	

26.	E		(1998 PI  Q8)
27.	B		(1992 PI  Q12)

Buoyancy/Upthrust

28.	A		(1997 PI  Q10)
29.	D		(1999 PI  Q6)
30.	E		(1992 PI  Q6)
31.	All forces on the balloon are balanced [N1].
	The weight of the balloon is equal and opposite
	to the upthrust force.

32a.			(1997 PI  Q32a)
	
  b.	0N		(1997 PI  Q32b)

33a.i.			(1996 PI  Q32ai)
	
  a.ii. Fbuoyancy = 1250N
  b. 	Fbuoyancy = 1250N (buoyancy force not dependent on depth)
34a.	w = 49000N		(1995 PII  Q4a)
  b.i.	Buoyancy = 20000N	(1995 PII  Q4bi/ii)
  b.ii.				
  c.	No effect		(1995 PII  Q4c)
35a.i.				(1994 PII  Q2)
	
  a.ii.	Buoyancy = 5650N
  a.iii.T = 750N
  b.	Trope 1 = 413.8N
	Trope 2 = 413.8N

  c.	Cold air is more dense than hot air and for a given 
	volume has a greater mass than hot air. Exchanging hot
	air for cold air increases the mass and weight of the 
	balloon. The weight of the balloon will now be greater 
	than the upthrust. This unbalanced force would result in 
	a downward acceleration.

36a.				(1999 PII  Q4a)
	 
  b.i.				(1999 PII  Q4bi)
  b.ii.	mass = 14790Kg
  b.iii.			(1999 PII  Q4biii)

Temperature Scales

37.	  0oC  = 273K
	 20oC  = 293K
	100oC  = 373K
	-50oC  = 223K
	-173oC = 100K

38.	300K = 27oC
	400K = 127oC
	650K = 377oC
	173K = -100oC
	4K   = -269oC
39.	A		(1993 PI  Q9)

40a.	77K		(1994 PI  Q33a)
  b.			(1994 PI  Q33b)

Pressure and Temperature

41a.	The pressure of a gas is directly proportional to its
	kelvin temperature.
	[P1/T1 = P2/T2]

  b.	Gas pressure is a result of collisions between gas 
	particles and the walls of their container. As the 
	temperature of the gas increases the kinetic energy 
	of the particles increases. The faster moving particles 
	collide more frequently, and forcefully, with the walls 
	of their container. The increase in average force per unit 
	area of container wall is the reason for the pressure
	increase.  
	 	
42.

43.	-273oC is called absolute zero. It is the coldest 
	possible temperature.

44.	P = 120kPa
45.	T = 750K (477oC)
46.	C			(1998 PI  Q9)
47.	E			(1994 PI  Q10)
48.	D			(1999 PI  Q10)

49a.i.						(1992 PII  Q4ai)

 
 a.ii.	Pressure/Temperature(K) = Constant	(1992 PII  Q4aii)
  a.iii.					(1992 PII  Q4aiii)
  b.	T = 68.3K				(1992 PII  Q4b)

Volume and Temperature

50.	T = 546K(273oC)
51a.	
  b.	V = 0.14cm3
  c.	Plot a graph of volume against kelvin
	temperature for the known temperatures of 
	boiling water and melting ice.
	Record the volume for the freezing mixture
	and use the the graph to estimate the kelvin
	temperature of the mixture.

	Or

	Use the relationship:
	V1/T1 = V2/T2
	And solve for the unknown temperature(T2) by substituting 
	in known values of V1,T1,V2 and solving for T2.
	
  d.	The relationship: V1/T1 = V2/T2, assumes the gas is
	at constant pressure, equal to atmospheric pressure. 
	A change in atmospheric pressure would change the gas pressure. 
	A graph, or constant(P/T), would therefore be required for each 
	pressure of the gas.

Pressure and Volume

52.	E		(1992 PI  Q11)
53.	A   		(1995 PI  Q10)

54.	When a gas is compressed into a smaller volume 
	the particles collide more frequently with the
	walls of their container. A geater average 
	force is therefore exerted by the gas on a 
	smaller area. This is equivalent to an increase
	in pressure. 

55.	V = 1500(litres)
56.	V = 56ml

57a.	V = 1800(litres)
  b.	900 litres available at 200kPa
	time = available volume/volume per minute

	t = 900/25
	t = 36(minutes)

58. 	P = 4x105Pa
59.	V2 = 3.84m3		(1995 PI  Q33)
60a.				(1996 PII  Q4a)

  
  b.	P = 295kPa (using average value of the constant PV)(1996 PII  Q4b)
  c.							   (1996 PII  Q4c)
  d.							   (1996 PII  Q4d)

General Questions

61.	D		(1999 PI  Q9)
62.	A		(1997 PI  Q11)

63a.	m = 3.87x10-3kg		(1998 PII  Q4a)
  b.i.	F = 2100N		(1998 PII  Q4bi)
  b.ii.	P2 = 1.5x105Pa		(1998 PII  Q4bii)

64a.	V2 = 1600cm3			(1993 PII  Q4a)
  b.	The pressure increases.		(1993 PII  Q4b)
  	The volume remains constant.
	The temperature increases.

  c.	T2 = 750K			(1993 PII  Q4c)