# Mass and weight of air in a room

Given:

A problem in my thermodynamics text is stated as follows...

Determine the mass and the weight of the air contained in a room whose dimensions are $V=$ $15ft$ x $20ft$ x $20ft$. Assume the density of the air is $\rho=0.0724\cdot\frac{lbm}{ft^3}$.

My Solution:

First find the mass...

$$m=\rho\times V$$

$$m=0.0724\cdot\frac{lbm}{ft^3}\times 6000\cdot ft^3$$

$$=434.3\cdot lbm$$

Now find the force acting on the air due to gravity. This is the weight of the air assumed at sea-level...

$$W=m\times g$$

$$W=434.3\cdot lbm\times32.174\cdot\frac{ft}{s^2}$$

$$=13976\cdot lbf$$

Question:

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

• For extra fun: atmospheric air is $\sim 14.7 psi$, so the total force on the floor of the room (20 x 20) is more than 800,000 pounds!
– Dan
Apr 4 '15 at 0:57

A pound force is defined as the force required to accelerate a slug at 1 ft/s^2. The density of air is $\rho = 0.0724 \ lb_m/ft^3 = 0.0724/32.2 \ slugs/ft^3$
The weight of the air is $\rho V g = 0.0724/32.2 \ slugs/ft^3 \cdot32.2 ft/s^2\cdot 6000 ft^3 = 0.0724\cdot 6000 \ slugs\ ft/s^2 = 434.4 lb_f$