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...

$$W=m\times g$$

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

$$=13976\cdot lbf$$


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. thermodynamics push
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edited Apr 6 "15 at 23:15

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asked Apr 3 "15 at 23:18

Jules MansonJules Manson
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1 answer 1

energetic earliest Votes
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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