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...

$$W=m\times g$$

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

$$=13976\cdot lbf$$


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.

  • 2
    $\begingroup$ 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! $\endgroup$
    – Dan
    Commented Apr 4, 2015 at 0:57

1 Answer 1


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 $\rho V g = 0.0724/32.2 \ slugs/ft^3 \cdot32.2 ft/s^2\cdot 6000 ft^3 = 0.0724\cdot 6000 \ slugs\ ft/s^2 = 434.4 lb_f$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.