Chapter 3, Problem 3A.6E

(a)

Interpretation Introduction

Interpretation:

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ is height difference in $\text{Torr}$ has to be determined.

Concept Introduction:

Force exerted by gas on unit area is pressure. It is can be calculated as follows:

  $P=\frac{F}{A}$

Here,

$\text{P}$ is pressure, $\text{F}$ is force and $\text{A}$ is area of an object.

Pressure of an atmosphere can be measured through manometer by employing the relation as follows:

  $P=dhg$

Here,

$d$ is density

$h$ is height of mercury.

$g$ is acceleration of free fall.

Manometer is a device as barometer employed to measure pressure of gas trapped in container.

Answer to Problem 3A.6E

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ isheight difference in $\text{Torr}$ is $243.1822\text{Torr}$.

Explanation of Solution

Pressure of an atmosphere measured through manometer is given by relation as follows:

  $P=dhg$        (1)

Substitute $24.32\text{cm}$ for $h$, $13595\text{ kg}\cdot {\text{m}}^{-3}$ for $d$ and $9.806\text{m}\cdot {\text{s}}^{-2}$ for $g$ in equation (1) to calculate $P$ in $\text{Torr}$.

  $\begin{array}{c}P=\left(13595\text{ kg}\cdot {\text{m}}^{-3}\right)\left(24.32\text{cm}\right)\left(\frac{0.01\text{m}}{\text{1}\text{cm}}\right)\left(9.806\text{m}\cdot {\text{s}}^{-2}\right)\\ =32421.617\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}\left(\frac{1\text{Pa}}{\text{1}\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}}\right)\left(\frac{\text{0}\text{.00750062}\text{Torr}}{1\text{Pa}}\right)\\ =243.1822\text{Torr}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ is height difference in $\text{atm}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3A.6E

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ isheight difference in $\text{atm}$ is $0.31997\text{atm}$.

Explanation of Solution

Substitute $24.32\text{cm}$ for $h$, $13595\text{ kg}\cdot {\text{m}}^{-3}$ for $d$ and $9.806\text{m}\cdot {\text{s}}^{-2}$ for $g$ in equation (1) to calculate $P$ in $\text{atm}$.

  $\begin{array}{c}P=\left(13595\text{ kg}\cdot {\text{m}}^{-3}\right)\left(24.32\text{cm}\right)\left(\frac{0.01\text{m}}{\text{1}\text{cm}}\right)\left(9.806\text{m}\cdot {\text{s}}^{-2}\right)\\ =32421.617\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}\left(\frac{1\text{Pa}}{\text{1}\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}}\right)\left(\frac{\text{9}\text{.8692}\times {\text{10}}^{-6}\text{atm}}{1\text{Pa}}\right)\\ =0.31997\text{atm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Pressure in $\text{Pa}$ inside apparatus at the end of the reaction when height difference is $24.32\text{cm}$ is has to be determined.

Concept Introduction:

Force exerted by gas on unit area is pressure. It is can be calculated as follows:

  $P=\frac{F}{A}$

Here,

$\text{P}$ is pressure, $\text{F}$ is force and $\text{A}$ is area of an object.

Pressure of an atmosphere can be measured through manometer by employing the relation as follows:

  $P=dhg$

Here,

$d$ is density

$h$ is height of mercury.

$g$ is acceleration of free fall.

Manometer is a device as barometer employed to measure pressure of gas trapped in container.

Answer to Problem 3A.6E

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ is height difference in $\text{Pa}$ is $1.3\times {10}^4\text{Pa}$.

Explanation of Solution

Substitute $24.32\text{cm}$ for $h$, $13595\text{ kg}\cdot {\text{m}}^{-3}$ for $d$ and $9.806\text{m}\cdot {\text{s}}^{-2}$ for $g$ in equation (1) to calculate $P$ in $\text{Pa}$.

  $\begin{array}{c}P=\left(13595\text{ kg}\cdot {\text{m}}^{-3}\right)\left(24.32\text{cm}\right)\left(\frac{0.01\text{m}}{\text{1}\text{cm}}\right)\left(9.806\text{m}\cdot {\text{s}}^{-2}\right)\\ =32421.617\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}\left(\frac{1\text{Pa}}{\text{1}\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}}\right)\\ =32421.617\text{Pa}\\ \approx 3.2421\times {10}^4\text{Pa}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ is height difference in $\text{bar}$ has to be determined.

Concept Introduction:

Force exerted by gas on unit area is pressure. It is can be calculated as follows:

  $P=\frac{F}{A}$

Here,

$\text{P}$ is pressure, $\text{F}$ is force and $\text{A}$ is area of an object.

Pressure of an atmosphere can be measured through manometer by employing the relation as follows:

  $P=dhg$

Here,

$d$ is density

$h$ is height of mercury.

$g$ is acceleration of free fall.

Manometer is a device as barometer employed to measure pressure of gas trapped in container.

Answer to Problem 3A.6E

Pressure inside apparatus at the end of the reaction when $24.32\text{cm}$ is height difference in $\text{bar}$ is $0.32421\text{bar}$.

Explanation of Solution

Substitute $24.32\text{cm}$ for $h$, $13595\text{ kg}\cdot {\text{m}}^{-3}$ for $d$ and $9.806\text{m}\cdot {\text{s}}^{-2}$ for $g$ in equation (1) to calculate $P$ in $\text{Bar}$.

  $\begin{array}{c}P=\left(13595\text{ kg}\cdot {\text{m}}^{-3}\right)\left(24.32\text{cm}\right)\left(\frac{0.01\text{m}}{\text{1}\text{cm}}\right)\left(9.806\text{m}\cdot {\text{s}}^{-2}\right)\\ =32421.617\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}\left(\frac{1\text{Pa}}{\text{1}\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}}\right)\left(\frac{{\text{10}}^5\text{bar}}{1\text{Pa}}\right)\\ =0.32421\text{bar}\end{array}$