91Ó°ÊÓ

At room temperature, the vapor pressure of mercury is \({P_{{\rm{vm}}}} = 0.0015\;{\rm{mm - Hg}}\).

At sea level, the height of mercury in a barometer is \(h = 760\;{\rm{mm}}\).

The atmospheric pressure is the sum of the pressure of the mercury's height and the mercury's vapor pressure.

Here, you have to use the equation,\({\bf{error = }}\frac{{{P_{{\rm{vapor}}}}}}{{{P_{{\rm{atm}}}}}}{\bf{ \times 100\% }}\).

The pressure of the mercury is \({P_{\rm{m}}} = \rho gh\), where \(\rho \) is the density of the mercury.

Now, the atmospheric pressure is:

\(\begin{aligned}{l}{P_{{\rm{atm}}}} &= {P_{{\rm{vm}}}} + {P_{\rm{m}}}\\{P_{\rm{m}}} &= {P_{{\rm{atm}}}} - {P_{{\rm{vm}}}}\end{aligned}\).

Therefore, you get:

\({P_{\rm{m}}} < {P_{{\rm{atm}}}}\)

Hence, the true atmospheric pressure is greater than the value read from the barometer.

The pressure of mercury is:

\(\begin{aligned}{c}{P_{\rm{m}}} &= \rho gh\\ &= \rho g \times 760\;{\rm{mm}}\\ &= 760\;{\rm{mm - Hg}}\end{aligned}\)

The error obtained is:

\(\begin{aligned}{c}{\rm{\% error}} &= \left( {\frac{{{P_{\rm{m}}} - {P_{{\rm{atm}}}}}}{{{P_{{\rm{atm}}}}}}} \right) \times 100\% \\ &= \left( {\frac{{ - {P_{{\rm{vm}}}}}}{{{P_{{\rm{atm}}}}}}} \right) \times 100\% \\ &= \left( {\frac{{ - 0.0015\;{\rm{mm - Hg}}}}{{760\;{\rm{mm - Hg}}}}} \right) \times 100\% \\ &= - 2.0 \times {10^{ - 4}}\% \end{aligned}\)

Hence, the percent error is \( - 2.0 \times {10^{ - 4}}\% \) for mercury.

The pressure of water vapor at STP is \({P_{{\rm{vw}}}} = 611\;{\rm{Pa}}\).

Now, the atmospheric pressure at STP is \({P_{{\rm{atm}}}} = 1.013 \times {10^5}\;{\rm{Pa}}\).

Therefore, the percent error is:

\(\begin{aligned}{c}{\rm{\% error}} &= \left( {\frac{{{P_{{\rm{vw}}}}}}{{{P_{{\rm{atm}}}}}}} \right) \times 100\% \\ &= \left( {\frac{{611\;{\rm{Pa}}}}{{1.013 \times {{10}^5}\;{\rm{Pa}}}}} \right) \times 100\% \\ &= 0.603\% \end{aligned}\)

Hence, the percent error for the water is \(0.603\% \).