In one observation, the column in a mercury barometer (as is shown in Figure) has a measured heighth of $\mathbf{740}\mathbf{.}\mathbf{35}\mathbf{}\mathit{m}\mathit{m}$. The temperature is$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{°}C$, at which temperature the density of mercury role="math" localid="1657193277146" $\mathit{\rho }$is $\mathbf{1}\mathbf{.}\mathbf{3608}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$. The free-fall acceleration g at the site of the barometer is $\mathbf{9}\mathbf{.}\mathbf{7835}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. What is the atmospheric pressure at that site in Pascal and in torr (which is the common unit for barometer readings)?

Measured height$h=740.35\mathrm{mm}$

Density of mercury,$\mathrm{\rho }=1.3608\times {10}^4\mathrm{kg}/{\mathrm{m}}^3$

Free fall acceleration at site of barometer, $g=9.7835\mathrm{m}/{\mathrm{s}}^2$

To find pressure, we can use the formula in terms of density, acceleration, and height. We can convert the pressure from Pascal to torr using the conversion factor.

Formula:

Pressure applied on a body, (i)

$1\mathrm{torr}=133.33\mathrm{Pa}$

Atmospheric pressure in Pascal using equation (i) is given as:

$$\begin{gathered} p=1.3608\times {10}^4\mathrm{kg}/{\mathrm{m}}^3\times 9.7835\mathrm{m}/{\mathrm{s}}^{2}x0.74035\mathrm{m} \\ =98565.66\mathrm{Pa}.....................................\left(\mathrm{a}\right) \end{gathered}$$

Hence, the value of pressure in Pascal is$98565.66Pa$

From equation (a), we get the pressure, hence

$$\begin{gathered} p=98565.66\mathrm{Pa}\times \frac{1\mathrm{torr}}{133.33\mathrm{Pa}}(\because 1\mathrm{torr}=133.33\mathrm{Pa}) \\ =739.26\mathrm{torr} \end{gathered}$$

Hence, the value of pressure in torr is role="math" localid="1657194537605" $739.26\mathrm{torr}$.