SM0486

difference of the two volumes) and 0.0012 gm/cm3 (the density of air), we determine

the value (in terms of mass) of the error caused by the buoyancy of air is

k. We could also figure the buoyant effect air has on each of the two

different masses and then subtract to find the difference in buoyant effect. Since

the brass standard had the greatest volume, we can state that air had the greatest

effect on it.

From the preceding example since the brass was our standard, we

should add the mass correction due to buoyancy to the stainless steel mass.

l. Now, using the information and examples, determine the buoyancy correction

for the 100-gram stainless steel mass in the example after we arbitrarily change

its density from 10 grams per cm3 to 12.5 grams per cm3. Use the questions which

follow as a guide to your solution:

(1) What value or values do you need?

Examine the example.

The volumes of the

unknown and the standard were computed.

Since brass is our standard in this

problem, its volume is the same. For the stainless steel mass with the arbitrary

density of 12.5 grams per cm3 its volume becomes:

(2) What is the second step, after the volumes for stainless steel and

brass are known? As stated before, one easy method of finding an equivalent mass

for buoyancy is to find the difference between the two masses and then find the

product of the calculated differences and the density of air, (*ΔV*) (). This means

that your next step is to determine the differences in volume for the 100-gram mass

standard brass and the 100-gram mass of stainless steel.

(3) You know the difference between the two volumes.

What next?

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