Question

(III) When using a mercury barometer (Section 10–6), the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height h of mercury in a barometer is about 760 mm. (a) If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? (b) What is the percent error? (c) What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?

Short answer

(a) The true atmospheric pressure is greater than the value read from the barometer.

(b) The percent error is \( - 2.0 \times {10^{ - 4}}\% \) for mercury.

(c) The percent error for the water is \(0.603\% \).

Step-by-step solution

Given data

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 \(h = 760\;{\rm{mm}}\).

Concepts

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,\({\bf{error = }}\frac{{{P_{{\rm{vapor}}}}}}{{{P_{{\rm{atm}}}}}}{\bf{ \times 100\% }}\).

Explanation for part (a)

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.

Calculation of part (b)

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.

Calculation of part (c)

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 \(0.603\% \).