Chapter 5, Problem 42E

(a)

Interpretation Introduction

Interpretation:

The pressure of the gases in given two situations of manometers (a) and (b) (Exercise-41) containing non-volatile silicone oil should be determined in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ if the manometer shows a reading of 118mm and 215mm respectively, and the advantages of using less dense fluid than mercury in a manometer used to measure relatively small differences in pressure should be explained.

Concept Introduction:

• The manometer is a devise used measure the pressure of a gas. The pressure of gas is determined by the value of ‘h’ shown by the manometer. This ‘h’-value is added or subtracted with atmospheric pressure to determine the pressure of gas.
• The fluid used in a manometer can be varied.
• The pressure is directly proportional to the mass of fluid, because pressure is the force acting per area.
• Equation for density is,

       $\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

• The mass is directly proportional to the volume of a fluid.

Pressure of a substance can be stated in various units. The units of pressure are interconvertible. The relations between units of pressure are,

• Since the unit $\text{mmHg}$ and the unit $\text{torr}$ is used interchangeably.

$1\text{mm}\text{Hg}\text{=}1\text{torr}$

• Conversion of 1 $\text{torr}$ into $\text{atm}$ is,

                     $\text{1}\text{mm}\text{Hg}\text{=}\frac{\text{1}}{\text{760}}\text{atm }$

• The 1 $\text{mmHg}$ pressure in $\text{pascals}$ unit is,

                                 $\text{1}\text{mm}\text{Hg}\text{=}\frac{\text{101325}}{\text{760}}\text{Pa}$

To find: the pressure of the gases in given two situations of manometers (a) and (b) (Exercise-41) containing non-volatile silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ when the manometer shows a reading of 118mm and 215mm respectively and explain the advantages of using less dense fluid than mercury in a manometer used to measure relatively small differences in pressure.

Answer to Problem 42E

The pressure of the gas in manometer (a) containing silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ are,

$\text{749torr}$, $\text{0}\text{.986atm}$, $\text{9}\text{.99}\times {\text{10}}^{\text{4}}\text{Pa}$

The pressure of the gas in given situation of manometer (b) containing silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ are,

$\text{781torr}$, $\text{1}\text{.03atm}$, $\text{1}\text{.04}\times {\text{10}}^5\text{Pa}$

Explanation of Solution

The relation between pressures measured in the manometers containing mercury and silicone oil, if the volumes are same for manometers.

The density of silicon oil is gives as $\text{1}\text{.3}{\text{g/cm}}^{\text{3}}$.

The density of mercury is gives as $\text{13}\text{.6}{\text{g/cm}}^{\text{3}}$.

Equation for density is,

       $\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

The mass is directly proportional to the volume of a fluid.

The manometers containing mercury and silicone oil are having same volumes of fluid.

Therefore,

${\text{V}}_{\text{Hg}}\text{=}{\text{V}}_{\text{Si}}\text{}\text{=}\frac{{\text{m}}_{\text{Hg}}}{{\text{d}}_{\text{Hg}}}\text{}\text{}\text{=}\frac{{\text{m}}_{\text{Si}}}{{\text{d}}_{\text{Si}}}$

The pressure is directly proportional to the mass of a fluid, because pressure is the force acting per area.

Hence, the above equation becomes,

$\frac{{\text{P}}_{\text{Hg}}}{{\text{d}}_{\text{Hg}}}\text{}\text{}\text{=}\frac{{\text{P}}_{\text{Si}}}{{\text{d}}_{\text{Si}}}$

Therefore,

The pressures in mercury manometer is,

$\begin{array}{l}{\text{P}}_{\text{Si}}\text{=}\frac{{\text{d}}_{\text{Si}}}{{\text{d}}_{\text{Hg}}}\times {\text{P}}_{\text{Hg}}=\frac{\text{1}\text{.3}{\text{g/cm}}^{\text{3}}}{\text{13}\text{.6}{\text{g/cm}}^{\text{3}}}\times {\text{P}}_{\text{Hg}}\\ =0.0956\times {\text{P}}_{\text{Hg}}\end{array}$

Hence,

The relation between pressures measured in the manometers containing mercury and silicone oil is,

${\text{P}}_{\text{Si}}\text{}\text{=}\text{(0}\text{.0956}\text{}{\text{)P}}_{\text{Hg}}$

Convert the pressures measured in the manometers containing mercury (Exercise-41) in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ to pressures measured in the manometers containing silicone oil .

The pressure of the gas in manometer (a) containing silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ are,

$\text{749torr}$, $\text{0}\text{.986atm}$, $\text{9}\text{.99}\times {\text{10}}^{\text{4}}\text{Pa}$

The pressure of the gas in given situation of manometer (b) containing silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ are,

$\text{781torr}$, $\text{1}\text{.03atm}$, $\text{1}\text{.04}\times {\text{10}}^5\text{Pa}$

From the Exercise-41,

• The pressure of the given gas (figure-a) in mercury manometer is =   $\text{(760}-118\text{)}\text{mm}\text{Hg}$

• The pressure of the given gas (figure-b) in mercury manometer is = $\text{(760}+118\text{)}\text{mm}\text{Hg}$

The conversion formula for manometers containing mercury to manometers containing silicone oil is,

${\text{P}}_{\text{Si}}\text{}\text{=}\text{(0}\text{.0956}\text{}{\text{)P}}_{\text{Hg}}$, this equation is applicable to column of manometer (h value), because the equation is derived from the volumes of columns of manometers containing mercury and silicon oil.

Therefore,

• The pressure of the given gas (figure-a) in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ by using silicon oil manometer is,

 =   $\text{(760}-0.956\times 118\text{)}\text{mm}\text{Hg}$

=   $749\text{mm}\text{Hg}$

The calculated pressure is $749\text{mm}\text{Hg}$; the $\text{mmHg}$ and $\text{torr}$ units are used interchangeably,

Therefore,

$749\text{mm}\text{Hg}\text{=}749\text{torr}$

The calculated pressure is $749\text{mm}\text{Hg}$. So the pressure in $\text{atm}$ unit is,

$\text{749}\text{mm}\text{Hg}\text{=}\text{749}\text{mm}\text{Hg}\times \frac{\text{1}}{\text{760}\text{mm}\text{Hg}}\text{atm }$

     =      $\text{0}\text{.986}\text{atm}$

The calculated pressure is $749\text{mm}\text{Hg}$. So the pressure in $\text{pascals}$ unit is,

$\text{749}\text{mm}\text{Hg}\text{=}\text{749}\text{mm}\text{Hg}\times \frac{\text{101325}}{\text{760}\text{mm}\text{Hg}}\text{Pa}$

   $\text{=}\text{9}\text{.99}\times {\text{10}}^4\text{Pa}$

• The pressure of the given gas (figure-b) in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ by using silicon oil manometer is,

=   $\text{(760}+0.956\times 118\text{)}\text{mm}\text{Hg}$

=   $781\text{mm}\text{Hg}$

The calculated pressure is $781\text{mm}\text{Hg}$; the $\text{mmHg}$ and $\text{torr}$ units are used interchangeably,

Therefore,

$781\text{mm}\text{Hg}\text{=}781\text{torr}$

The calculated pressure is $781\text{mm}\text{Hg}$. So the pressure in $\text{atm}$ unit is,

$\text{781}\text{mm}\text{Hg}\text{=}781\text{mm}\text{Hg}\text{*}\frac{\text{1}}{\text{760}\text{mm}\text{Hg}}\text{atm }$

     =      $\text{1}\text{.03}\text{atm}$

The calculated pressure is $781\text{mm}\text{Hg}$. So the pressure in $\text{pascals}$ unit is,

$781\text{mm}\text{Hg}\text{=}781\text{mm}\text{Hg}\text{*}\frac{\text{101325}}{\text{760}\text{mm}\text{Hg}}\text{Pa}$

   $\text{=}\text{1}\text{.04}\times {\text{10}}^5\text{Pa}$

Hence,

The pressure of the gas in manometer (a) containing silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ are,

$\text{749torr}$, $\text{0}\text{.986atm}$, $\text{9}\text{.99}\times {\text{10}}^{\text{4}}\text{Pa}$

The pressure of the gas in given situation of manometer (b) containing silicone oil in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ are,

$\text{781torr}$, $\text{1}\text{.03atm}$, $\text{1}\text{.04}\times {\text{10}}^5\text{Pa}$

Conclusion

The pressure of the gases in given two situations of manometers containing mercury in figure (a) and (b) in units of $\text{mmHg}$ are known according to In Exercise-41, so the pressure of the gases in given two situations of manometers containing silicon oil in figure (a) and (b) in units of $\text{mmHg}$ are determined by using the conversion formula of different density fluids. These determined $\text{mmHg}$ units of pressures are converted to $\text{torr}$, $\text{atm}$ and $\text{pascals}$ units by using the corresponding unit conversional formulas.

The volume of silicone oil or the column height (h-value) of the silicone oil manometer is higher than that of column height (h-value) of mercury manometer. To measure relatively small differences in pressure, the column height should be high. Therefore, the less dense fluid will advantage for measuring relatively small differences in pressure

(b)

Interpretation Introduction

Interpretation:

The pressure of the gases in given two situations of manometers (a) and (b) (Exercise-41) containing non-volatile silicone oil should be determined in units of $\text{torr}$, $\text{atm}$ and $\text{pascals}$ if the manometer shows a reading of 118mm and 215mm respectively, and the advantages of using less dense fluid than mercury in a manometer used to measure relatively small differences in pressure should be explained.

Concept Introduction:

• The manometer is a devise used measure the pressure of a gas. The pressure of gas is determined by the value of ‘h’ shown by the manometer. This ‘h’-value is added or subtracted with atmospheric pressure to determine the pressure of gas.
• The fluid used in a manometer can be varied.
• The pressure is directly proportional to the mass of fluid, because pressure is the force acting per area.
• Equation for density is,

       $\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

• The mass is directly proportional to the volume of a fluid.

Pressure of a substance can be stated in various units. The units of pressure are interconvertible. The relations between units of pressure are,

• Since the unit $\text{mmHg}$ and the unit $\text{torr}$ is used interchangeably.

$1\text{mm}\text{Hg}\text{=}1\text{torr}$

• Conversion of 1 $\text{torr}$ into $\text{atm}$ is,

                     $\text{1}\text{mm}\text{Hg}\text{=}\frac{\text{1}}{\text{760}}\text{atm }$

• The 1 $\text{mmHg}$ pressure in $\text{pascals}$ unit is,

                                 $\text{1}\text{mm}\text{Hg}\text{=}\frac{\text{101325}}{\text{760}}\text{Pa}$

Answer to Problem 42E

The measurement will be more precise.

Explanation of Solution

Explain the advantages of using less dense fluid than mercury in a manometer used to measure relatively small differences in pressure should be explained.

The measurement will be more precise.

Take the less dense fluid as compare to mercury is as silicon oil.

The density of silicon oil is gives as $\text{1}\text{.3}{\text{g/cm}}^{\text{3}}$.

The density of mercury is gives as $\text{13}\text{.6}{\text{g/cm}}^{\text{3}}$.

Equation for density is,

       $\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

From the density equation, the density of a fluid is inversely proportional to volume of the fluid.

That means,

$\frac{{\text{V}}_{\text{Si}}}{{\text{V}}_{\text{Hg}}}\text{}\text{=}\frac{{\text{d}}_{\text{Hg}}}{{\text{d}}_{\text{Si}}}\text{}\text{}$

Therefore,

$\begin{array}{l}{\text{V}}_{\text{Si}}\text{}\text{=}\frac{{\text{d}}_{\text{Hg}}}{{\text{d}}_{\text{Si}}}\text{}\text{}\times {\text{V}}_{\text{Hg}}=\frac{13.6}{1.3}\times {\text{V}}_{\text{Hg}}\\ =10.5\times {\text{V}}_{\text{Hg}}\end{array}$

The volume of silicone oil or the column height (h-value) of the silicone oil manometer is 10.5 times higher than that of column height (h-value) of mercury manometer.

To measure relatively small differences in pressure, the column height should be high.

Therefore, the less dense fluid will advantage for measuring relatively small differences in pressure.

Conclusion

The pressure of the gases in given two situations of manometers containing mercury in figure (a) and (b) in units of $\text{mmHg}$ are known according to In Exercise-41, so the pressure of the gases in given two situations of manometers containing silicon oil in figure (a) and (b) in units of $\text{mmHg}$ are determined by using the conversion formula of different density fluids. These determined $\text{mmHg}$ units of pressures are converted to $\text{torr}$, $\text{atm}$ and $\text{pascals}$ units by using the corresponding unit conversional formulas.

The volume of silicone oil or the column height (h-value) of the silicone oil manometer is higher than that of column height (h-value) of mercury manometer. To measure relatively small differences in pressure, the column height should be high. Therefore, the less dense fluid will advantage for measuring relatively small differences in pressure