Ok so I just read an article that said that 3000 tons of sea water was contamina

Question

Ok so I just read an article that said that 3000 tons of sea water was contaminated in Fukushima. So I wanted to know the volume of 3000 tones of seawater; to figure out the oceans percent contamination.

Average temp in japan 29C or 302.15K=T

Vapor pressure of salt water at 29C = 30.08 Torr = 0.039578953 ATM=P

R=0.082057 L*ATM/K*mol

MW(H2O)=18.01528 g/mol

PV = (nRT) ==> P / R T = n / V ===> P / (RT) = (g / MW ) / V ===> (P*MW) / (RT) =  g / L

g/L= [ P(atm)*MW(H2O) ] / [ R(.082075 L*ATM/ K* mol) * T(k) ]

g/L = 0.02875855655298  this is my density used for my conversion

3000 tonnes ===> (2,205 lbs / 1 tonnes) (3000 tonnes) = 6.615 E 6 lbs

6.615E6 lbs===> (453.59237 g / lb ) (6.615E6 lbs) = 3.000513528 E 9 grams

(3.000513528 E 9 g) (1/(0.02875855655298 g/L ) ) = 1.043346359 E 11 L contaminated

Liters of h20 in the ocean 1.26 E 21 L

(1.043346359 E 11 contaminated / 1.26 E 21 L Ocean total ) (100 for percent) = 8.28 E -9 % contaminated.

My question is did I screw up somewhere? Thanks for your help in advance.

Explanation / Answer

You have taken "vapour pressure" of sea water.....which is pressure exerted by vapour on water

Hencne this vapour is actually GAS and then you have taken this gas pressure to calculate gas density

BUT actually You wanted to KNOW THE VOLUME OF WATER and no the volume of VAPOUR OF WATER (GAS)

You could have simply used the relation density = mass/volume

ANyway, great effort......at least you applied science in pratical

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