Chapter 5, Problem 4QAP

Interpretation Introduction

(a)

Interpretation:

The pressure 42.7 kPa is to be converted into mmHg, atm and psi.

Concept introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP

The pressure in psi is 6.20 psi

The pressure in mm Hg is $3.20\times {10}^2\text{ mm Hg}$.

The pressure in atm is 0.422 atm.

Explanation of Solution

The force exerted by gas over a unit surface of area is called pressure of gas. The expression of pressure is as below:

$P\text{ =}\frac{F}{A}$

Where, P is pressure of gas, F is force exerted by gas and A is area.

The SI unit of pressure is Pascal, which is denoted as Pa. It is represented as follows;

$1{\text{ Pa = 1 kgm}}^{-1}{S}^{-2}$

The pressure given in kPa is 47.7 kPa. The pressure is also represented in psi, that is, pound-force square inch. The value of one Pascal is equivalent to $1.45\times {10}^{-4}psi$. The pressure is converted into psi as follows:

$\begin{array}{l}P\text{ = 42}\text{.7 kPa}\\ \text{ =(42}\text{.7 kPa)}\left(\frac{{10}^3\text{ Pa}}{1\text{ kPa}}\right)\left(\frac{1.45\times {10}^{-4}psi}{1\text{ Pa}}\right)\\ \text{ =6}\text{.20 psi}\end{array}$

Therefore, the pressure in psi is 6.20 psi.

The value of one Pascal pressure is equivalent to 9.87 atm. The pressure is converted into atmosphere as follow:

$\begin{array}{l}P\text{ = 42}\text{.7 kPa}\\ \text{ =(42}\text{.7 kPa)}\left(\frac{{10}^3\text{ Pa}}{1\text{ kPa}}\right)\left(\frac{9.87\times {10}^{-6}\text{ atm}}{1\text{ Pa}}\right)\\ \text{ =0}\text{.422 atm}\end{array}$

Hence, the pressure in atm is 0.422 atm.

The value of one atmosphere is equivalent to 760 mm Hg. The pressure is converted into mm Hg as follow:

$\begin{array}{l}P\text{ = 0}\text{.422 atm}\\ \text{ =(0}\text{.422 atm)}\left(\frac{760\text{ mm Hg}}{1\text{ atm}}\right)\\ \text{ =3}\text{.20 }\times {\text{10}}^2\text{ mm Hg}\end{array}$

Hence, the pressure in mm Hg is $3.20\times {10}^2\text{ mm Hg}$.

Interpretation Introduction

(b)

Interpretation:

The pressure 29.1 psi is to be converted into mmHg, atm and kPa.

Concept introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP

The pressure in kPa is 201 kPa.

The pressure in mm Hg is $1.50\times {10}^3\text{ mm Hg}$.

The pressure in atmosphere is 1.98 atm.

Explanation of Solution

The pressure given in pound-force square inch is 29.1 psi. The value of one psi pressure is equivalent to 6.90kPa. The pressure is converted into kPa as follows:

$\begin{array}{l}P\text{ = }29.1psi\\ \text{ =(}29.1psi\text{)}\left(\frac{6.90\text{ kPa}}{1\text{ psi}}\right)\\ \text{ =}201\text{ kPa}\end{array}$

Hence, the pressure in kPa is 201 kPa.

The value of one Pascal pressure is equivalent to 9.87 atm. The pressure is converted into atmosphere as follows:

$\begin{array}{l}P\text{ = }42.7\text{ kPa}\\ \text{ =(}201\text{ kPa)}\left(\frac{{10}^3\text{ Pa}}{1\text{ kPa}}\right)\left(\frac{9.87\times {10}^{-6}\text{ atm}}{1\text{ Pa}}\right)\\ \text{ =1}\text{.98 atm}\end{array}$

Hence, the pressure in atmosphere is 1.98 atm.

The value of one atmosphere is equivalent to 760 mmHg. The pressure is converted into mm Hg as follows:

$\begin{array}{l}P\text{ = 1}\text{.98 atm}\\ \text{ =(1}\text{.98 atm)}\left(\frac{760\text{ mm Hg}}{1\text{ atm}}\right)\\ \text{ =1}\text{.50}\times {\text{10}}^3\text{ mm Hg}\end{array}$

Hence, the pressure in mm Hg is:

$1.50\times {10}^3\text{ mm Hg}$

Interpretation Introduction

(c)

Interpretation:

The pressure 0.788 atm is to be converted into mmHg, psi and kPa.

Concept introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP

The pressure in mm Hg is 599 mm Hg.

The pressure in psi is 11.6 psi.

The pressure in kPa is 79.8 kPa.

Explanation of Solution

The value of one atmosphere is equivalent to 760 mmHg. The 0.788 atmosphere pressure is converted into mm Hg as follows:

$\begin{array}{l}P\text{ = 0}\text{.788 atm}\\ \text{ =(0}\text{.788 atm)}\left(\frac{760\text{ mm Hg}}{1\text{ atm}}\right)\\ \text{ =}599\text{ mm Hg}\end{array}$

Hence, the pressure in mm Hg is 599 mm Hg.

The value of one atmosphere pressure is equivalent to 101.33 kPa. The pressure is converted into kPa as follows:

$\begin{array}{l}P\text{ = 0}\text{.788 atm}\\ \text{ =(0}\text{.788 atm)}\left(\frac{101.33\text{ kPa}}{1\text{ atm}}\right)\\ \text{ =}79.8\text{ kPa}\end{array}$

Hence, the pressure in kPa is 79.8 kPa.

The value of one Pascal is equivalent to $1.45\times {10}^{-4}$ psi. The pressure is converted into psi as follows:

$\begin{array}{l}P\text{ = }79.8\text{ kPa}\\ \text{ =(}79.8\text{ kPa)}\left(\frac{{\text{10}}^3\text{ kPa}}{1\text{ kPa}}\right)\left(\frac{1.45\times {10}^{-4}\text{ psi}}{1\text{ Pa}}\right)\\ \text{ =}11.6\text{ psi}\end{array}$

Hence, the pressure in psi is 11.6 psi.

Interpretation Introduction

(d)

Interpretation:

The pressure 1216 mm Hg is to be converted into atm, psi, and kPa.

Concept introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP

The pressure in atmosphere is 1.60 atm.

The pressure in psi is 23.5 psi.

The pressure in kPa is 162.1 kPa

Explanation of Solution

The given pressure value is 1216 mm Hg. The value of one atmosphere is equivalent to 760 mmHg. The pressure is converted into atmosphere as follows:

$\begin{array}{l}P\text{ = 1216 mm Hg}\\ \text{ =(1216 mm Hg)}\left(\frac{\text{1 atm}}{760\text{ mm Hg}}\right)\\ \text{ =}1.60\text{ atm}\end{array}$

Therefore, the pressure in atmosphere is 1.60 atm.

The value of one atmosphere pressure is equivalent to 101.33 kPa. The pressure is converted into kPa as follows:

$\begin{array}{l}P\text{ = }1.60\text{ atm}\\ \text{ =(}1.60\text{ atm)}\left(\frac{\text{101}\text{.33 kPa}}{1atm}\right)\\ \text{ =}162.1\text{ kPa}\end{array}$

Hence, the pressure in kPa is 162.1 kPa.

The value of one Pascal is equivalent to

$1.45\times {10}^{-4}$ psi. The pressure is converted into psi as follows:

$\begin{array}{l}P\text{ = }162.1\text{ kPa}\\ \text{ =(}162.1\text{ kPa)}\left(\frac{{\text{10}}^3\text{ Pa}}{1\text{ kPa}}\right)\left(\frac{1.45\times {10}^{-4}\text{ psi}}{1\text{ Pa}}\right)\\ \text{ =23}\text{.5 psi}\end{array}$

Hence, the pressure in psi is 23.5 psi.