Determine the mass and also the weight of the air consisted of in a room whose dimensions room $V=$ $15ft$ x $20ft$ x $20ft$. Assume the density of the air is $\rho=0.0724\cdot\fraclbmft^3$.

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**My Solution:**

First uncover the mass...

$$m=\rho\times V$$

$$m=0.0724\cdot\fraclbmft^3\times 6000\cdot ft^3$$

$$=434.3\cdot lbm$$

Now uncover the pressure acting top top the air due to gravity. This is the load of the waiting assumed in ~ sea-level...

$$W=m\times g$$

$$W=434.3\cdot lbm\times32.174\cdot\fracfts^2$$

$$=13976\cdot lbf$$

**Question:**

I find it hard to believe that in an median size room the air weighs a whopping $14,000\cdot lbf$. Did I perform something wrong in my calculations or is this correct? If this is correct maybe we earthlings living on the surface ar of the earth are the actual *extremophiles*.

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edited Apr 6 "15 at 23:15

706Astor

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asked Apr 3 "15 at 23:18

Jules MansonJules Manson

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A pound force is characterized as the force required to accelerate a cheese at 1 ft/s^2.The thickness of wait is $\rho = 0.0724 \ lb_m/ft^3 = 0.0724/32.2 \ slugs/ft^3$

The weight of the waiting is $\rho V g = 0.0724/32.2 \ slugs/ft^3 \cdot32.2 ft/s^2\cdot 6000 ft^3 = 0.0724\cdot 6000 \ slugs\ ft/s^2 = 434.4 lb_f$

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answered Apr 4 "15 at 6:50

Ronny LandsverkRonny Landsverk

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