from Anne FrankHow high would the level be in an alcohol barometer at a pressure of 0.72atm? The density of alcohol is 0.79×103kg/m3, and 1atm=1.013×105n/m2?
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To determine how high the level would be in an alcohol barometer at a pressure of 0.72 atm, we can use the barometric formula which relates the pressure exerted by a column of fluid to the height of the fluid column, the density of the fluid, and the acceleration due to gravity.
The formula for the pressure exerted by a fluid column is:
\[ P = h \cdot \rho \cdot g \]
where: - \( P \) is the pressure exerted by the fluid column (in Pascal, Pa) - \( h \) is the height of the fluid column (in meters, m) - \( \rho \) is the density of the fluid (in kilograms per cubic meter, kg/m³) - \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth)
Rearrange the formula to solve for the height (\( h \)):
\[ h = \frac{P}{\rho \cdot g} \]
You've provided the density of alcohol as \( \rho = 0.79 \times 10^{3} \) kg/m³ and the atmospheric pressure conversion factor 1 atm = \( 1.013 \times 10^{5} \) N/m².
To find the height of the alcohol column at a pressure of 0.72 atm, first convert the pressure to Pascals:
\[ P = 0.72 \text{ atm} \times \frac{1.013 \times 10^{5} \text{ N/m}^2}{1 \text{ atm}} = 0.72 \times 1.013 \times 10^{5} \text{ N/m}^2 \]
\[ P = 7.2936 \times 10^{4} \text{ N/m}^2 \]
Now plug in the values for \( P \), \( \rho \), and \( g \) into the height formula:
\[ h = \frac{7.2936 \times 10^{4} \text{ N/m}^2}{0.79 \times 10^{3} \text{ kg/m}^3 \times 9.81 \text{ m/s}^2} \]
\[ h = \frac{7.2936 \times 10^{4}}{0.79 \times 10^{3} \times 9.81} \text{ m} \]
\[ h \approx \frac{7.2936 \times 10^{4}}{7749} \text{ m} \]
\[ h \approx 9.41 \text{ m} \]
Therefore, the height of the alcohol column in the barometer at a pressure of 0.72 atm would be approximately 9.41 meters.