  Question

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  B

150 mm  C

125 mm  D

133.3 mm  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=1.9625×10−4 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) - 4993.63×1.9625×10−4=3745.22×Vf Vf=2.6166×10−4 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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