Piston Pressure and Unit Conversion

[Nim R]

Problem:
CO₂ is contained in a vertical cylinder at a pressure of 30 atm by a piston with a mass of 57 lb_m. If g is 32.4 ft/sec² and the barometric pressure is 29.7 in. Hg what is the area of the piston?

Relevant equations
F = Ma/g_c
W = Mg/g_c
P= F/A
g_c = 32.174 lb_m ft / (lb_f sec²)
1 atm = 29.9 in. Hg
1 atm = 2116.217 lb_f / ft²

Attempt (4 sig figs)
1) 29.7 in. Hg * 1 atm / (29.9 in. Hg) = 0.9933 atm
2) 30 atm - 0.9933 atm = 29.01 atm * 2116.217 lb_f / ft² / (1 atm) = 61,380 lb_f / ft²
3) 61,380 lb_f / ft² = (57 lb_m) * (32.4 ft / sec²) / (A * (32.174 lb_m ft / (lb_f sec²))
4) A = 9.352 x 10^-4 ft² = 0.1347 in²

I feel like this number is way too small. Was the assumption that the force of the piston was only its weight on the gas wrong?

Thanks

 

[TomHart]

I'm pretty rusty on these types of problems, but if I take 60 lb and apply it to an area of 0.001 ft² (which is 0.144 in²), I get a pressure of 60000 lb/ft². Your number certainly appears to be in the ballpark.

 

[Nim R]

Well good to know my answer numerically makes sense. Intuitively though I don't have a clue what something with those dimensions and strength would look like.

Thanks for the reply!