Question

A tall cylinder with a cross-sectional area 13.0 cm^2 is partially filled with mercury; the surface of the mercury is 8.50 cm above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

Answer 1

Answer:

Approximately $1.5\times 10^{3}\; {\rm cm^{3}}$ (approximately $1.5 \; {\rm L}$.)

Explanation:

Look up the density of mercury: $\rho({\rm Hg}) \approx 13.5\; {\rm g\cdot cm^{-3}}$.

The density of water is $\rho(\text{water}) = 1.000\; {\rm g \cdot cm^{-3}}$.

Multiply the horizontal cross-section area $A$ by height $h$ to find the volume of mercury in this cylinder:

$\begin{aligned}V({\rm Hg}) &= h({\rm Hg})\, A \end{aligned}$.

The mass of that much mercury will be:

$\begin{aligned} m({\rm Hg}) &= V({\rm Hg})\, \rho({\rm Hg}) \\ &= h({\rm Hg}) \, A\, \rho({\rm Hg})\end{aligned}$.

The mass of the liquid is proportional to the pressure that the liquid exerts on the bottom of the cylinder.

In this question, the pressure of the added water need to match that of the mercury in the container, so that the total pressure will double. Hence, the mass of the added water will need to be equal to that of the mercury in the cylinder.

$m(\text{water}) = m(\text{Hg})$.

$V(\text{water})\, \rho(\text{water}) = V({\rm Hg})\, \rho({\rm Hg})$.

$\begin{aligned} V(\text{water}) &= \frac{V({\rm Hg})\, \rho({\rm Hg})}{\rho(\text{water})} \\&= \frac{h({\rm Hg})\, A\, \rho({\rm Hg})}{\rho(\text{water})} \\ &= \frac{8.50\; {\rm cm} \times 13.0\; {\rm cm^{2}} \times 13.5\; {\rm g \cdot cm^{-3}}}{1.000\; {\rm g \cdot cm^{-3}}} \\ &\approx 1.5\times 10^{3}\; {\rm cm^{3}} \\ &\approx 1.5 \; {\rm L}\end{aligned}$.