WEIGHT HANGING FROM A PISTON
Statement
In a vertical cylinder, of 10 cm in diameter and 0.5 m of height, open on the bottom, a 10 kg piston with
its top surface at 20 cm from the cylinder closed end, encloses a certain amount of carbon dioxide. Find:
a) The equilibrium pressure.
b) The amount of gas trapped.
En un cilindro vertical de 10 cm de diámetro y 0,5 m de altura, abierto por abajo, un émbolo de 10 kg
cuya cara superior está a 20 cm del fondo del cilindro, encierra una cierta cantidad de dióxido de carbono.
Determinar
a) La presión en el equilibrio.
b) La cantidad de gas atrapado.
Solution
a) The equilibrium pressure.
First a sketch to help visualise the system:
Fig. 1. Sketch of a cylinder with a hanging piston, and the piston free-body force diagram.
Assuming that friction between piston and cylinder is not dominant; what prevents the piston
falling down out of the cylinder? Suction of the trapped gas? Vacuum? Can pressure be negative?
A main advantage of science is that, once the basic Principles of Nature have been mastered,
applying the fundamental laws solves the problem with no hesitation; the force balance on the
piston as a free body at rest, gives the answer:
 pA  m g  p A 
p p
 F  0 
P
0
0

mP g
10  9.8
 105 
 87.5 kPa
2
D
 0.12
4
4
where a standard value of 100 kPa for the ambient pressure is assumed, in lack of more precise
data.
In practise, friction would never be totally negligible, and, if such an experiment were tried in the
lab (e.g. with a tight-fitting self-lubricating graphite piston in a precision-machined cylinder), a
range of acceptable values might be 87..88 kPa. And do not forget that the less friction, the more
difficult to keep the piston seal air-tight (a low-friction piston will eventually fall down the
cylinder).
Weight hanging from a piston
1
b) The amount of gas trapped.
From pV=nRT, n=pAz/(RT)=87.5·103·(0.12/4)/(8.3·288)=0.058 mol (2.5 g), where a standard
value of 288 K (15 ºC) for the ambient temperature is assumed, in lack of more precise data.
Comments
A lesson to learn
The author regrets to point out the sad fact that, having proposed such an exercise to some 100 entryuniversity-students, only some 20% of them managed to solve the first part of this problem (accepting as
valid such offending nonsense as p=87555.282 Pa!). There were many types of mistakes, from which we
(all, students and teachers) should learn a good lesson:
1. Some 30% were unaware of the force balance, and simply answered p=100 kPa at equilibrium,
or looked for a solution through some astray relation like pV=constant (or worse, reasoning
2.
3.
4.
5.
that "if gases at STP-conditions occupy 22.4 litres, to the given 1.6 litres of CO2 should
correspond a pressure of 100·1.6/22.4=7 kPa").
Some 30% made 'just a sign mistake' and answered p=p0+mg/A=112 kPa.
Some 20% did not pay attention to atmospheric pressure (perhaps induced by the generalised
used of manometric pressure in most ordinary-life instances (e.g. tyre pressure, hydrostatic
pressure, and so on).
Some 20% made mistakes on geometrical calculations, setting e.g. V=Dz, V=Dz, V=D2z, or
A=D2, or on trivial unit conversions like setting 10 cm=0.01 m.
For the second question, some 20% used the expression m=V with the density taken from a
tabulation of liquid data.
6. For the second question, some 10% used the expression m=RT/(pV), in spite that it was an
open-book exercise.
7. For the second question, some 10% made mistakes using the gas constant, mixing-up units.
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Weight hanging from a piston
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