Example 9.2: Calculation of Barometric Pressure

Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury \( = Failed to interpret math QQQ 13.6 g/cm^3.QQQ\)

Solution

\(h_{\mathrm{\ce{Hg},column}}\) \(= 760.\ \mathrm{mm}\)


\(ρ_{\mathrm{\ce{Hg}}}\) \(= 13.6\ \frac{\mathrm{g}}{\mathrm{cm}^{3}}\)


\(P_{\mathrm{hydrostatic}} = h_{\mathrm{column}} \cdot ρ \cdot g_{\mathrm{0}}\)     

\(g_{\mathrm{0}}\) \(= 9.81\ \frac{\mathrm{m}}{\mathrm{s}^{2}}\)


\(P_{\mathrm{hydrostatic}}\) \(= h_{\mathrm{\ce{Hg},column}} \cdot ρ_{\mathrm{\ce{Hg}}} \cdot g_{\mathrm{0}}\)

\(\ \ \ =760.\ \mathrm{mm} \cdot 13.6\ \frac{\mathrm{g}}{\mathrm{cm}^{3}} \cdot 9.81\ \frac{\mathrm{m}}{\mathrm{s}^{2}}\)

\(\ \ \ =10.336\ \frac{\mathrm{g}}{\mathrm{mm}^{2}} \cdot 9.81\ \frac{\mathrm{m}}{\mathrm{s}^{2}}\)

\(\ \ \ =1.014\times 10^{5}\ \frac{\mathrm{g}}{\mathrm{mm}\ \mathrm{s}^{2}}\)


\(P_{\mathrm{hydrostatic}}\) \(= 1.014\times 10^{5}\ \frac{\mathrm{g}}{\mathrm{mm}\ \mathrm{s}^{2}}\)

\(\ \ \ =1.001\ \mathrm{atm}\)


The calculation supports the claim.