Chapter 5, Problem 20QAP

Interpretation Introduction

(a)

Interpretation:

To use the ideal gas law to complete the blank columns of the given table for propane gas.

Concept introduction:

The ideal gas law states the relationship between the pressure, volume, number of moles and temperature of gas at ideal conditions and it is calculated using the following formula:

$PV=nRT$

Here P is the pressure of the system, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature.

Mole is a ratio between mass and molar mass. It can be calculated as follows:

$n=\frac{\text{mass}}{\text{molar mass}}$

Answer to Problem 20QAP

The bank columns in each row are filled in bold. The completed table is as follows:

Sub part	Pressure	Volume	Temperature	Moles	Grams
a	18.9psi	0.886L	220C	0.047	2.1

Explanation of Solution

As per the given table we have to find the moles and grams of propane gas. First let us calculate the molar mass of $C_3{H}_8$ :

$\begin{array}{l}{\text{Molar mass of C}}_{\text{2}}{\text{O}}_{\text{4}}\text{=2×molar mass of carbon+4× molar mass of oxygen}\\ \text{=2×12 g/mol+4×16 g/mol}\\ \text{=44 g/mol}\end{array}$

We have to convert psi into atm using the following formula:

$\begin{array}{l}\text{14}\text{.7 psi=1 atm}\\ \text{18}\text{.9 psi=}\frac{\text{1}}{\text{14}\text{.7}}\text{×18}\text{.9 atm}\\ \text{=1}\text{.29 atm}\end{array}$

Let us use the ideal gas equation to calculate the number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ :

$PV=nRT$

$\begin{array}{l}\text{1}\text{.29 atm×0}\text{.886 L=n×0}\text{.0821 L}\text{.atm/mol}\text{.K×295 K}\\ \text{n=}\frac{\text{1}\text{.29 atm×0}\text{.886 L}}{\text{(0}\text{.0821 L}\text{.atm/mol}\text{.K)×(295 K)}}\\ \text{=0}\text{.047 mol}\end{array}$

Now let us substitute these values in number of moles formula to find the mass in grams:

$\begin{array}{l}\text{0}\text{.047 mol=}\frac{\text{mass}}{\text{44 g/mol}}\\ \text{mass=0}\text{.047 mol×44 g/mol}\\ \text{=2}\text{.1 g}\end{array}$

Interpretation Introduction

(b)

Interpretation:

To use the ideal gas law to complete the blank columns of the given table for propane gas.

Concept introduction:

The ideal gas law states the relationship between the pressure, volume, number of moles and temperature of gas at ideal conditions and it is calculated using the following formula:

$PV=nRT$

Here P is the pressure of the system, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature. We know that mole is a ratio between mass and molar mass therefore let us substitute this in the ideal gas equation to obtain density:

Mole is a ratio between mass and molar mass. It can be calculated as follows:

$n=\frac{\text{mass}}{\text{molar mass}}$

Answer to Problem 20QAP

The bank columns in each row are filled in bold. The completed table is as follows:

Sub part	Pressure	Volume	Temperature	Moles	Grams
b	633mm Hg	1.993L	-330C	0.0844	3.72

Explanation of Solution

As per the given table, we have to find the temperature and grams of propane gas. In sub part a, we have found that the MM is 44g/mol and as given, the number of moles is 0.0844mol therefore let us substitute these values in number of moles formula:

$\begin{array}{l}\text{0}\text{.0844 mol=}\frac{\text{mass}}{\text{44g/mol}}\\ \text{mass=0}\text{.0844 mol×44 g/mol}\\ \text{=3}\text{.72 g}\end{array}$

Now we must convert mmHg into atm using the following formula:

$\begin{array}{l}\text{760 mmHg=1 atm}\\ \text{633 mmHg=}\frac{\text{1}}{\text{760}}\text{×633 atm}\\ \text{=0}\text{.83 atm}\end{array}$

Let us use the ideal gas equation to calculate the temperature of $C_3{H}_8$ :

$PV=nRT$

$\begin{array}{l}\text{0}\text{.83 atm×1}\text{.993 L=0}\text{.0844 mol×0}\text{.0821 L}\text{.atm/mol}\text{.K×T}\\ \text{T=}\frac{\text{0}\text{.83 atm×1}\text{.993 L}}{\text{0}\text{.0844 mol×0}\text{.0821 L}\text{.atm/mol}\text{.k}}\text{=240 K}\\ {\text{=-33}}^{\text{0}}\text{C}\end{array}$

Interpretation Introduction

(c)

Interpretation:

To use the ideal gas law to complete the blank columns of the given table for propane gas.

Concept introduction:

The ideal gas law states the relationship between the pressure, volume, number of moles and temperature of gas at ideal conditions and it is calculated using the following formula:

$PV=nRT$

Here P is the pressure of the system, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature. We know that mole is a ratio between mass and molar mass therefore let us substitute this in the ideal gas equation to obtain density:

Mole is a ratio between mass and molar mass. It can be calculated as follows:

$n=\frac{\text{mass}}{\text{molar mass}}$

Answer to Problem 20QAP

The bank columns in each row are filled in bold. The completed table is as follows:

Sub part	Pressure	Volume	Temperature	Moles	Grams
c	1.876atm	47.3 L	750C	2.842 mol	125.04 g

Explanation of Solution

As per the given table, we have to find the volume and grams of propane gas. First we have to convert ⁰F into K using the following formula:

$\begin{array}{l}Kelvin={(}^{0}F-32)\times \frac{5}{9}+273.15\\ =(75-32)\times \frac{5}{9}+273.15\\ =43\times \frac{5}{9}+273.15=380.65K\end{array}$

Let us use the ideal gas equation to calculate the volume of $C_3{H}_8$ :

$\begin{array}{l}\text{1}\text{.876 atm×V=2}\text{.842 mol×0}\text{.0821 L}\text{.atm/mol}\text{.K×380}\text{.65K}\\ \text{V=}\frac{\text{2}\text{.842 mol×0}\text{.0821 L}\text{.atm/mol}\text{.K×380}\text{.65 K}}{\text{1}\text{.876 atm}}\\ \text{=47}\text{.3 L}\end{array}$

Now let us substitute these values in number of moles formula to find the mass in grams:

$\begin{array}{l}\text{2}\text{.842 mol=}\frac{\text{mass}}{\text{44 g/mol}}\\ \text{mass=2}\text{.842 mol×44 g/mol}\\ \text{=125}\text{.048 g}\end{array}$

Interpretation Introduction

(d)

Interpretation:

To use the ideal gas law to complete the blank columns of the given table for propane gas.

Concept introduction:

The ideal gas law states the relationship between the pressure, volume, number of moles and temperature of gas at ideal conditions and it is calculated using the following formula:

$PV=nRT$

Here P is the pressure of the system, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature. We know that mole is a ratio between mass and molar mass therefore let us substitute this in the ideal gas equation to obtain density:

Mole is a ratio between mass and molar mass. It can be calculated as follows:

$n=\frac{\text{Mass}}{\text{Molarmass}}$

Answer to Problem 20QAP

The bank columns in each row are filled in bold. The completed table is as follows:

Sub part	Pressure	Volume	Temperature	Moles	Grams
d	11.2atm	2244mL	130C	1.07	47.25

Explanation of Solution

The pressure and moles of propane gas needs to be calculated. As per the given information substitute 47.25 g/mol for mass and 44 g/mol for MM in the number of moles formula:

$\begin{array}{l}n=\frac{\text{47}\text{.25 g}}{\text{44 g/mol}}\\ \text{=1}\text{.07 mol}\end{array}$

Let us use the ideal gas equation to calculate the pressure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ :

$\begin{array}{l}\text{P×2}\text{.244 L=1}\text{.07 mol×0}\text{.0821 L}\text{.atm/mol}\text{.K×286 K}\\ \text{P=}\frac{\text{1}\text{.07 mol×0}\text{.0821 L}\text{.atm/mol}\text{.k×286 K}}{\text{2}\text{.244 L}}\\ \text{=11}\text{.2 atm}\end{array}$