If a barometer were built using water $\left(d=1.0 \mathrm{g} / \mathrm{cm}^{3}\right)\( instead of mercury \)\left(d=13.6 \mathrm{g} / \mathrm{cm}^{3}\right)$ , would the column of water be higher than, lower than, or the same as the column of mercury at 1.00 atm ? If the level is different, by what factor? Explain.

Short Answer

Expert verified
A barometer built with water instead of mercury would have a higher water column at 1.00 atm. The water column would be around 10.33 m, while the mercury column would be approximately 0.76 m. The height of the water column would be 13.6 times higher than the height of the mercury column.

Step by step solution

01

Determine the hydrostatic pressure for both columns

Hydrostatic pressure can be calculated using the formula: \(P = \rho gh \) Where \(P\) is the pressure, \(\rho\) is the density of the liquid, \(g\) is the acceleration due to gravity, and \(h\) is the height of the liquid column. For water: Density (\(\rho_w\)) = 1.00 g/cm³ For mercury: Density (\(\rho_m\)) = 13.6 g/cm³ At 1.00 atm, the pressure (\(P\)) = 1.00 × 101325 Pa = 1013.25 hPa (1 m = 100 cm)
02

Find the height of the water column

Using the formula for hydrostatic pressure, we solve for the height of the water column (\(h_w\)): \( P = \rho_w g h_w \) \( h_w = \frac{P}{\rho_w g} \) Converting the units for \(\rho_w\) into kg/m³: \( \rho_w = \frac{1000}{1000} \times 1000 \, kg/m³ = 1000 \, kg/m³\) Now, we can find the height of the water column: \( h_w = \frac{101325 \, Pa}{1000 \, kg/m³ \times 9.81 \, m/s²} \) \( h_w \approx 10.33\, m \)
03

Find the height of the mercury column

Using the same hydrostatic pressure formula, we solve for the height of the mercury column (\(h_m\)): \( P = \rho_m g h_m \) \( h_m = \frac{P}{\rho_m g} \) Converting the units for \(\rho_m\) into kg/m³: \( \rho_m = \frac{13.6}{1000} \times 1000 \times 1000 \times \frac{1}{1000} \, kg/m³ = 13600 \, kg/m³\) Now, we can find the height of the mercury column: \( h_m = \frac{101325 \, Pa}{13600 \, kg/m³ \times 9.81 \, m/s²} \) \( h_m \approx 0.76\, m \)
04

Compare the heights of the two columns and find the factor

Comparing the heights of the two columns, we find: \( h_w \approx 10.33\, m \) (Water column height) \( h_m \approx 0.76\, m \) (Mercury column height) The height of the water column is higher than the height of the mercury column, so the column of water would be higher at 1.00 atm. Now, let's find the factor by which the height of the water column differs from that of the mercury column: Factor = \(\frac{h_w}{h_m}\) Factor ≈ \(\frac{10.33 \, m}{0.76 \, m}\) Factor ≈ 13.6
05

Conclusion

A barometer built with water instead of mercury would have a column of water that is higher than the column of mercury, at 1.00 atm. The height of the water column would be 13.6 times higher than the height of the mercury column.

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