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

Answer 1

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.
$\therefore Downwardforce=4\times (0.25\times 9.8)N$
= 9.8 N.- - - - - - (1)
The upward force on the piston was provided by the pressure on the bottom face, given as -
$P_{initial}\times Areaofpiston$. - - - - - - (2)
Equating (1) and (2)
$P_{initial}\times \pi \left(\frac{50}{1000}\right)\times \left(\frac{50}{1000}\right)\times \left(\frac{1}{4}\right)=9.8$
$P_{initial}=4993.63Pa.$
The initial volume of the gas is -

$=\pi \left(\frac{d^2}{4}\right)\times n=\pi \times 0.05\times 0.4\times \left(\frac{1}{4}\right)m^3$
$\Rightarrow V_i=1.9625\times {10}^{-4}{m}^3$.
Since the temperature of the initial and final states are the same, from ideal gas equation, we get -
$P_i\times V_i=P_f\times V_f.$- - - - - - (3)
$P_f$ can be calculated in a very similar fashion as$P_i$, the difference being - only 3 balls need to be considered now -
$P_f=\frac{3\times 0.25\times .8}{\pi \times \frac{(0.05{)}^2}{4}}Pa$
=3745.22 Pa.

Plugging all the values in (3) -
$4993.63\times 1.9625\times {10}^{-4}=3745.22\times V_f$
$V_f=2.6166\times {10}^{-4}{m}^3.$
From here, we can find the height, h, using the formula $V=\pi \left(\frac{d^2}{4}\right)h.$
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 ${\left(\frac{3}{4}\right)}^{th}$ to 3 balls.
The pressure required to support these 3 balls will also become${\left(\frac{3}{4}\right)}^{th}$.

Since PV = constant, the new volume will become ${\left(\frac{4}{3}\right)}^{rd}$ the original volume.
Hence the new height of the cylinder will be ${\left(\frac{4}{3}\right)}^{rd}$ the original height = 133.3 mm!