Atmospheric mass...math problem

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Can anyone help me finish this math problem? I think I got it started...

The density of air at ordinary atmospheric pressure and 25 degree celsius is 1.19 g/L. What is the mass, in kilograms, of the air in the room that measures 12.5 x 15.5 x 8.0 feet?

First, I converted the room dimensions given in feet to inches and then, centimeters.

12.5 feet = (12.5)(12) inches = (12.5)(12)(2.54) cm =381 cm.
15.5 feet = (15.5)(12) inches = (15.5)(12)(2.54) cm = 472.44 cm
8.0 feet = (8.0) (12) inches = (8.0) (12) (2.54) cm = 243. 84 cm

Am I on the right track in solving the problem? How do I get to the answer in kilograms?


My sincere thanks to anyone who can help me.

You only have to multiplie 381 cm x 472.44 cm x 243. 84 cm = 43891112cm³ for the volumen
/ 10cm³ = 438911,12 L and than change it liters.

10cm³ are a liter

438911,12 L * 1,19g/L = 522304,23 g =522,30423 kg than you multiplie your liter with your liter per grain g/L and than you change your grain to kilo grain, since kilo being latin for thousand you get one kg for thousand grain..

Just remember 1 dm^3 = 1 L. By far the easiest way to avoid all confusion with multipliers in more dimensions is simply inputting in correct unit - in your example 381 cm = 38,1 dm ... so volume of the room is 38,1x47,244x24,384 L = 43891 L and then multiply it by the density to get the mass which comes out roughly 52 kg.

But in case you do want to sort corrent units at the end just try like this. You have for example volume in cm^3, to convert it to liters (which equals dm^3) just input 1 dm = 10 cm so 1 L = 1 dm^3 = (10 cm)^3 = 10^3 cm^3 = 1000 cm^3.