Calculate the height in meters of a column of liquid glycerol (d = 1.26 g/cm³)required to exert the same pressure as 760 mmHg (d = 13.60 g/cm^3). A)8.20 m B)8.20 × 10³ m C)70.4 m D)0.704 m E)44.4 m

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Quizplus · Accepted Answer

To find the height of the glycerol column, we use the principle that the pressure exerted by a column of liquid is given by $P = h \cdot d \cdot g$, where $P$ is the pressure, $h$ is the height of the liquid column, $d$ is the density of the liquid, and $g$ is the acceleration due to gravity. Since the pressure exerted by the glycerol column is equal to the pressure exerted by a 760 mm column of mercury, we can equate the two pressures.Given:- Density of glycerol ($d_{glycerol}$) = 1.26 g/cm³ = 1260 kg/m³ (since 1 g/cm³ = 1000 kg/m³)- Density of mercury ($d_{Hg}$) = 13.60 g/cm³ = 13600 kg/m³- Height of mercury column ($h_{Hg}$) = 760 mm = 0.76 m (since 1000 mm = 1 m)- $g$ (acceleration due to gravity) is approximately 9.81 m/s², but it cancels out in the calculation since it's the same for both liquids.The pressure exerted by the mercury column is $P_{Hg} = h_{Hg} \cdot d_{Hg} \cdot g$.The pressure exerted by the glycerol column is $P_{glycerol} = h_{glycerol} \cdot d_{glycerol} \cdot g$.Setting $P_{Hg} = P_{glycerol}$ and solving for $h_{glycerol}$, we get:$$h_{glycerol} = \frac{h_{Hg} \cdot d_{Hg}}{d_{glycerol}}$$Substituting the given values:$$h_{glycerol} = \frac{0.76 m \cdot 13600 kg/m³}{1260 kg/m³}$$$$h_{glycerol} = \frac{10336}{1260}$$$$h_{glycerol} \approx 8.20 m$$Therefore, the correct answer is 8.20 m, which corresponds to choice A.

Calculate the Height in Meters of a Column of Liquid