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The Bat Life: The diameter of the cylinder in which the helium gas is stored is 50 mm and the weight of a steel ball is 0.25 kg. If initially there were 4 steel balls and the piston was 100 mm from the base, what would be the final position of the piston after dislodging 1 steel ball? Consider that the helium in the canister has the same temperature initially and at the end.



A

166.7 mm

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B

150 mm

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C

125 mm

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D

133.3 mm

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Solution

The correct option is D

133.3 mm


Since a very dilute form of Helium gas is stored in the canister, ideal gas equation is applicable -
PV=NkT.
Initially, the piston was pushed down by 4 steel balls.
 Downward force=4×(0.25×9.8)N
= 9.8 N.- - - - - - (1)
The upward force on the piston was provided by the pressure on the bottom face, given as -
Pinitial× Area of piston. - - - - - - (2)
Equating (1) and (2)
Pinitial×π(501000)×(501000)×(14)=9.8
Pinitial=4993.63 Pa.
The initial volume of the gas is -

=π(d24)×n=π×0.05×0.4×(14)m3
Vi=1.9625×104 m3.
Since the temperature of the initial and final states are the same, from ideal gas equation, we get -
Pi×Vi=Pf×Vf.- - - - - - (3)
Pf can be calculated in a very similar fashion asPi, the difference being - only 3 balls need to be considered now -
Pf=3×0.25×.8π×(0.05)24Pa
=3745.22 Pa.

Plugging all the values in (3) -
4993.63×1.9625×104=3745.22×Vf
Vf=2.6166×104 m3.
From here, we can find the height, h, using the formula V=π(d24)h.
The new height/position of the piston will be = 133.3 mm
Now that we have calculated the position by doing complex calculations, let's see if there is a more elegant solution.
In the initial state the gas supported 4 balls. In the final state that reduced by (34)th to 3 balls.
The pressure required to support these 3 balls will also become(34)th.

Since PV = constant, the new volume will become (43)rd  the original volume.
Hence the new height of the cylinder will be (43)rd the original height = 133.3 mm!
 

Physics

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