[rob000001]

Hey guys, Im having a hard time with this word problem. If anyone can solve pls. thx
The problem says:

The new safety standard proposed by EPA for micro particles in air is for particles up to 2.5 µmeter in diameter, the maximum allowable amount is 50µgram/ meter3 air. If your 10.0ft x 8.25ft x 12.5ft dorm room just meets the new EPA standard, how many particles are in your room? ( Assume particles are spherical and are primarily soot which has a density of 2.5g/cm3). Volume of a sphere = 4/3 π r3

[Borek]

Show how you tried.

[rob000001]

Show how you tried.

i tried to start by converting the ft into cm.. like 10.0ft*12in/1ft*1cm/2.54cm=47.2cm
^12.5ft=59.1cm
^8.25ft=39.0cm
then i multiplied them all together..=108791cm^3
then im supposed to use that into D=M/V i suppose, not really sure what to do next..

[DrCMS]

i tried to start by converting the ft into cm.. like 10.0ft*12in/1ft*1cm/2.54cm=47.2cm
^12.5ft=59.1cm
^8.25ft=39.0cm

Think about what you just wrote "10ft = 47.2cm"!!!!!!!!  What plant do you live on for Gods sake.  Here on earth 10ft = 305cm

[rob000001]

oh i converted wrong, here's what i meant..
10.0ft=305cm or 304.8cm
8.25ft= 251cm or 251.46cm
12.5ft=381cm
Multiply thru by rounded answ...=29167455 cm^3
then what do i do??

[Borek]

To use EPA standard you need room volume in m³ - either conver ft to m (before multiplication), or convert now cm³ to m³.

[rob000001]

so now it would be:
10.0ft=3.048m
8.25ft=2.5146m
12.5ft=3.81m
mult. thru..=29.2 m^3
now what do i do with this..

[Borek]

now what do i do with this..

You start thinking

You are ready to calculate mass of the soot in the room.

[rob000001]

This is what ive done so far::
Given::
2.5 umeter in diameter for particles
50 u gram/meter^3 max amount of particles
dorm room=10.ftx8.25ftx12.5ft-->cm^3 = 2.920148x10^7 cm^3.
V=4/3 pi*r^3
density = 2.5 grams/cm^3
so,
 v=(4/3) pi*(1.25)^3= 8.181231x10^-12 cm^3
----------------------------------------
V dorm room/ Vmax particles= 2.9201748x10^7 cm^3/ 8.181231x10^-12 cm^3= 3.56936x10^18 cm^3.(so this is the particles that fit in room)
----------------------------------------
50 ugrams/m^3 = 50 grams/cm^3

[D for density..]
Dmax/ Dprimary = 50 grams/cm^3/ 2.5grams/cm^3 = 20 -(20 times a particles fit in room)
so,
20(3.56936x10^18) = 7.13872x10^19 particles in room.

   Is this right??

[Borek]

V dorm room/ Vmax particles= 2.9201748x10^7 cm^3/ 8.181231x10^-12 cm^3= 3.56936x10^18 cm^3.(so this is the particles that fit in room)

That means no air is left in the room. It is all filled with particles.

50 ugrams/m^3 = 50 grams/cm^3

Completely wrong. 50 g/cm³ is 2.5 times more dense then depleted uranium - which is already pretty dense.

You have 29.2 cubic meter of air, each contains 50 micrograms of substance. How much substance you have?