My diving instructor claimed that the amount of energy required to fill a scuba

Question

My diving instructor claimed that the amount of energy required to fill a scuba tank is sufficient to lift a locomotive engine several feet in the air, which makes me take extra care with handling the things each time I take one off the rack. Calculate the energy needed to fill a standard scuba tank of volume 12.3 liters to a pressure of 200 atmospheres, assuming the filling pump operates gradually enough that the gas is maintained at the ambient temperature of the room. Explain the equations you use and justify their applicability. Then compare that result to the instructor's claim by calculating the height that much energy could lift a 150-ton locomotive.

Explanation / Answer

Consider a box. It is is h high and has a top and bottom area of A units.
It is full of gas at pressure p and the height h is variable so as h varies so p varies.
The outside pressure is Pa

Let's be simplistic and use Charles's law that states p*v is constant for a given volume of gas provided the temperature stays constant.
So for a given starting pressure P0 and volume V0 we can deduce any later pressure as p = P0*V0 / v

But since the volume is v=h*A we can reason that
p = P0*V0/(h*A)

Now if we let the gas expand just a teeny bit (a distance dh) the work done (dW) is the distance moved multiplied by the force applied
Force is (p - Pa) * A so
dW = ((P0*V0)/(h*A) - Pa) * A * dh
dW = (P0*V0/h - Pa*A) * dh

So now we want to sum all the dW for all the dh for a range of h from our initial pressure/volume until the pressure reaches Pa.
So the initial limit is based on v=h*A giving h = V0/A
and the closing limit is based on p = P0 * V0 / (h * A) so for p=Pa we get h = P0*V0/(Pa * A)

W = integrate((P0*V0/h - Pa*A) * dh) using limits above

W = P0*V0 * integrate(h-1*dh) - A * Pa * integrate(dh)

W = P0*V0 * [logeh] - A * Pa * [h]

substitute in the initial and final values for h

W = P0*V0 * (loge(P0*V0/(A * Pa)) - loge(V0/A)) - A * Pa * (P0*V0/(A * Pa) - V0/A)

since log(a)-log(b) = log(a/b) we can simplify this a bit

W = P0*V0 * (loge((P0*V0/(A*Pa))/(V0/A)) - A*Pa*(P0*V0/(A*Pa)-V0/A)

now it all comes together with basic algebra

W = P0*V0 * loge(P0/Pa) - P0*V0 + Pa*V0

W = P0*V0 * (loge(P0/Pa)-1) + Pa*V0

So for the given data, W=888,101 J

For raising a 15o ton vehicle work done= mgh=150000*9.81*h

So equationg both, h= 0.6 m

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