For an atmospheric pressure of 97.3 kPa (absolute), determine the heights of the fluid columns in barometers containing the following liquids:

(a) Mercury
(b) Water
(c) Ethyl alcohol

Calculate the heights, including the effect of vapor pressure, and compare the results with those obtained by neglecting vapor pressure:

(d) Mercury without vapor pressure
(e) Water without vapor pressure
(f) Ethyl alcohol without vapor pressure

(g) Do these results support the widespread use of mercury for barometers?

In the given question firstly we need to calculate the height of the pressure formula with in the vapor pressure, that includes the pressure, and its formula is:

[tex]p(atm)_{vp} = \gamma .h +P_{vp}[/tex]

[tex]\to \boxed{ h = \frac{p(atm)_{vp} - p_{vp}}{\gamma}}[/tex]

The vapour pressure where [tex]_{vp}[/tex] pays any attention. Next mercury measurement,Ethyl alcohol and water.

for mercury calculation:

a)

[tex]h_{Hg \ vp } =\frac{101 \times 10^3 \frac{N}{m^2}- 1.6 \times 10^{-1} \frac{N}{m^2}}{133 \times 10^3 \frac{N}{m^3}}[/tex]

[tex]=\frac{101000- 0.16 }{133000}\\\\ =\frac{100999.84}{133000}\\\\= 0.759[/tex]

In the next step we calculate water value:

b)

[tex]h_{H_{2}0 \ vp } =\frac{101 \times 10^3 \frac{N}{m^2}- 1.77 \times 10^{3} \frac{N}{m^2}}{9.80 \times 10^3 \frac{N}{m^3}}[/tex]

[tex]=\frac{101000- 1770}{9800}\\\\=\frac{99230}{9800}\\\\= 10.125[/tex]

In the last step, now we calculate the value of ethyl alchohol:

c)

[tex]h_{EA \ vp } =\frac{101 \times 10^3 \frac{N}{m^2}- 5.9 \times 10^{3} \frac{N}{m^2}}{7.74 \times 10^3 \frac{N}{m^3}}[/tex]

[tex]=\frac{101000 - 5900 }{7740}\\\\ =\frac{95100 }{7740}\\\\=12.286[/tex]

The following would be the precise measurement of the heights without

Pressure of the vapour:

[tex]a) h_{Hg} = \frac{101\times 10^3 \frac{N}{m^2}}{133\times 10^3 \frac{N}{m^3}} \\\\[/tex]

[tex]= \frac{101 000 }{133000 } \\\\= \frac{101}{133 } \\\\= 0.759[/tex]

[tex]\\\\b) h_{H_{2} o} = \frac{101\times 10^3 \frac{N}{m^2}}{9.80 \times 10^3 \frac{N}{m^3}}[/tex]

[tex]= \frac{101}{9.80 }\\\\=10.30[/tex]

[tex]\\\\c) h_{EA} = \frac{101\times 10^3 \frac{N}{m^2}}{7.74 \times 10^3 \frac{N}{m^3}}[/tex]

[tex]= \frac{101}{7.74 } \\\\=13.04[/tex]

It may draw the substantial inference from of the calculated values. For the very first time. Mercury indicators and vapour pressure effects are insignificant and necessary height is within acceptable values for the mercury barometer tube (imagine). The alcohol barometer tube 12 metres high!).