Chapter 9, Problem 77QAP

Interpretation Introduction

(a)

Interpretation:

The physical state (s) of water is (are) present in the flask needs to be identified.

Concept introduction:

If partial pressure of water vapor at the given temperature and volume is less than $P^{\ast }$, then all the water formed will be in the vapor state. And if the partial pressure of water vapor at the given temperature and volume is greater than $P^{\ast }$, then some of the water vapor condensed into liquid and some will be in vapor state.

Ideal gas law representation is given below.

$PV=nRT$   ..........(1)

Here, n is number of moles.

P represents pressure, V represents volume, T and R represents temperature and gas constant respectively.

Answer to Problem 77QAP

Two physical states, liquid and vapor are present in the flask.

Explanation of Solution

Given:

Volume = 10.0 L

Mass of hydrogen gas = 0.400 g

Mass of oxygen gas = 3.2 g

Reaction:

${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O}$

Atomic mass of hydrogen is 1 amu and oxygen is 16 amu. Therefore, molar mass of ${\text{H}}_{\text{2}}$ molecule is 2 g/mol and molar mass of ${\text{O}}_{\text{2}}$ molecule is 32 g/mol. Thus,

1 mole of ${\text{H}}_{\text{2}}$ gas $=2\text{g}$.

1 mole of ${\text{O}}_{\text{2}}$ gas $=32\text{g}$.

The given reaction in the question is given below:

${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O}$

In the above reaction, $2$ moles ${\text{H}}_{\text{2}}$ of gas reacts with $1$ mole of ${\text{O}}_{\text{2}}$ gas to form two moles of ${\text{H}}_{\text{2}}\text{O}$. In other words, $4$ g of ${\text{H}}_{\text{2}}$ gas reacts with $32$ g of ${\text{O}}_{\text{2}}$ gas to form $36$ g of ${\text{H}}_{\text{2}}\text{O}$. If $0.\text{4}00$ g of ${\text{H}}_{\text{2}}$ gas react with $3.20$ g of ${\text{O}}_{\text{2}}$ gas it will form $3.6\text{g}$ or $0.2$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Once ${\text{H}}_{\text{2}}\text{O}$ is formed in the $10.0\text{L}$ flask, it will cooled at $27°\text{C}$. At this temperature, the saturation pressure of water $P^{\ast }=26.7\text{torr}\left(\text{or 3559}\text{.708}\text{Pa}\right)$. If the partial pressure of water vapor at the given temperature and volume is less than $P^{\ast }$, then all the water formed will be in the vapor state. And if the partial pressure of water vapor at the given temperature and volume is greater than $P^{\ast }$, then some of the water vapor condensed into liquid and some will be in vapor state.

Given,

Volume ( V) $=10.0\text{L}$

Temperature ( T) $=27°\text{C}$

Calculate moles ( n) of ${\text{O}}_{\text{2}}$ as follows:

$\begin{array}{c}\text{Moles}\text{of}{\text{O}}_{\text{2}}\left(n\right)\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ =\frac{3.6}{16}\\ =0.2\text{mol}\end{array}$

The volume is converted from L to $m^3$ :

The temperature is converted into Kelvin as follows:

$\begin{array}{c}\text{ 27}°\text{C=27 +273}\\ =300\text{K}\end{array}$

Put all values in equation (1).

$\begin{array}{c}P\left(0.01{\text{m}}^3\right)=\left(0.2\text{mol}\right)\left(\text{8}\text{.31}\text{J/mol}\text{K}\right)\left(300\text{K}\right)\\ P=\frac{\left(0.2\cancel{\text{mol}}\right)\left(\text{8}\text{.31}\text{J/}\cancel{\text{mol}}\cancel{\text{K}}\right)\left(300\cancel{\text{K}}\right)}{\left(0.01{\text{m}}^3\right)}\\ P=\frac{498.6\text{J}}{0.01{\text{m}}^3}\\ \left(\text{1}{\text{J/m}}^{\text{3}}\text{=1}\text{Pa}\right)\end{array}$

$P=49860\text{Pa}$

Therefore, P value is greater than $P^{\ast }$ therefore some water vapor condensed into liquid and some will be in vapor. Hence, two physical state liquid and vapor is present in the flask.

Interpretation Introduction

(b)

Interpretation:

The final pressure in the flask is to be determined.

Concept introduction:

Ideal gas law representation is given below.

$PV=nRT$   ..........(1)

Here, n is number of moles.

P represents pressure, V represents volume, T and R represents temperature and gas constant respectively.

Answer to Problem 77QAP

Final pressure in the flask is $\left(P^{\ast }\right)=26.7\text{torr}\left(\text{or 3559}\text{.708}\text{Pa}\right)$

Explanation of Solution

Given:

Volume = 10.0 L

Mass of hydrogen gas = 0.400 g

Mass of oxygen gas = 3.2 g

Reaction:

${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O}$

Atomic mass of hydrogen is 1 amu and oxygen is 16 amu. Therefore, molar mass of ${\text{H}}_{\text{2}}$ is 2 and molar mass of ${\text{O}}_{\text{2}}$ is 32. Thus,

1 mole of ${\text{H}}_{\text{2}}$ gas $=2\text{g}$.

1 mole of ${\text{O}}_{\text{2}}$ gas $=32\text{g}$.

The given reaction in the question is given below:

${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O}$

In this reaction, $2$ moles of ${\text{H}}_{\text{2}}$ gas reacts with one mole of ${\text{O}}_{\text{2}}$ gas to form two moles of ${\text{H}}_{\text{2}}\text{O}$. In other words, $4$ g of ${\text{H}}_{\text{2}}$ gas reacts with $32$ g of ${\text{O}}_{\text{2}}$ gas to form $36$ g of ${\text{H}}_{\text{2}}\text{O}$. If $0.\text{4}00$ g of ${\text{H}}_{\text{2}}$ gas react with $3.20$ g of ${\text{O}}_{\text{2}}$ gas it will form $3.6\text{g}$ or $0.2$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Once ${\text{H}}_{\text{2}}\text{O}$ is formed in the $10.0\text{L}$ flask, it will cooled at $27°\text{C}$. At this temperature, saturation pressure of water $P^{\ast }=26.7\text{torr}\left(\text{or 3559}\text{.708}\text{Pa}\right)$. If partial pressure of water vapor at given temperature and volume is less than $P^{\ast }$, then all the water formed will be in vapor state. And if partial pressure of water vapor at given temperature and volume is greater than $P^{\ast }$, then some of the water vapor condensed into liquid and some will be in vapor state.

Calculate number of moles $\left(n\right)$, Substitute the value of $P^{\ast }$ in place of pressure in equation (1).

Given,

Volume ( V) $=10.0\text{L}$

Temperature ( T) $=27°\text{C}$

Number of moles n $=0.2$

The volume is converted from L to $m^3$ :

$\begin{array}{c}\text{Volume=10}\text{.0}\cancel{\text{L}}\times \left(\frac{1{\text{m}}^3}{1000\cancel{\text{L}}}\right)\\ =0.01{\text{m}}^3\end{array}$

The temperature is converted into Kelvin as follows:

$\begin{array}{c}\text{ 27}°\text{C=27 +273}\\ =300\text{K}\end{array}$

$\begin{array}{c}\left(3559.707{\text{N/m}}^{\text{2}}\right)\left(0.01{\text{m}}^3\right)=n\left(8.31\text{J/mol}\cancel{\text{K}}\right)\left(300\cancel{\text{K}}\right)\\ n=\frac{\left(3559.707\text{N/}\cancel{{\text{m}}^{\text{2}}}\right)\left(0.01\text{m}\cancel{{}^3}\right)}{8.31\text{J/mol}\times 300}\\ n=\frac{35.60\text{Nm}}{2493\text{J/mol}}\\ \left(1\text{}\text{Nm=1}\text{J}\right)\end{array}$

$n=0.0143\text{mol}$

Since,

$\begin{array}{c}\text{1 mol of water=18}\text{g}\\ \text{0}\text{.0143}\text{mol of water=18}\text{g}\left(\text{0}\text{.0143}\right)\\ =0.2574\text{g}\end{array}$

In previous case assuming that all water is in vapor phase. Exceed the saturation pressure some of the water will get condensed and remaining vapor will have just the pressure equal to the partial pressure, i.e. $P^{\ast }$.

Hence, the final pressure in the flask is $\left(P^{\ast }\right)=26.7\text{torr}\left(\text{or 3559}\text{.708}\text{Pa}\right)$

Interpretation Introduction

(c)

Interpretation:

The pressure in the flask is to be determined, if $\text{3}\text{.2}\text{g}$ of each gas is used.

Concept introduction:

Ideal gas law representation is given below.

$PV=nRT$   ..........(1)

Here, n is number of moles.

P represents pressure, V represents volume, T and R represents temperature and gas constant respectively.

Answer to Problem 77QAP

The pressure in the flask is $3.45\text{atm}$.

Explanation of Solution

Atomic mass of hydrogen is 1 amu and oxygen is 16 amu. Therefore, molar mass of ${\text{H}}_{\text{2}}$ is 2 and molar mass of ${\text{O}}_{\text{2}}$ is 32. Thus,

1 mole of ${\text{H}}_{\text{2}}$ gas $=2\text{g}$.

1 mole of ${\text{O}}_{\text{2}}$ gas $=32\text{g}$.

The given reaction in the question is given below:

${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O}$

In the above reaction, $2$ moles of ${\text{H}}_{\text{2}}$ gas react with $1$ mole of ${\text{O}}_{\text{2}}$ gas to form $2$ moles of ${\text{H}}_{\text{2}}\text{O}$. In other words, $4$ g of ${\text{H}}_{\text{2}}$ gas reacts with $32$ g of ${\text{O}}_{\text{2}}$ gas to form $36$ g of ${\text{H}}_{\text{2}}\text{O}$. If $0.\text{4}00$ g of ${\text{H}}_{\text{2}}$ gas react with $3.20$ g of ${\text{O}}_{\text{2}}$ gas it will form $3.6\text{g}$ or $0.2$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Once ${\text{H}}_{\text{2}}\text{O}$ is formed in the $10.0\text{L}$ flask, it will cooled at $27°\text{C}$. At this temperature, saturation pressure of water $P^{\ast }=26.7\text{torr}\left(\text{or 3559}\text{.708}\text{Pa}\right)$. If partial pressure of water vapor at given temperature and volume is less than $P^{\ast }$, then all the water formed will be in vapor state. And if partial pressure of water vapor at given temperature and volume is greater than $P^{\ast }$, then some of water vapor condensed into liquid and some will be in vapor state.

If $3.2\text{g}$ of each gas is used $2.8\text{g}$ or $1.4\text{moles}$ of ${\text{H}}_{\text{2}}$ gas will remain as it is because $0.\text{4}00$ g of ${\text{H}}_{\text{2}}$ is sufficient to react $3.20$ g of ${\text{O}}_{\text{2}}$.

Given,

Volume ( V) $=10.0\text{L}$

Temperature ( T) $=27°\text{C}$

Calculate number of moles ( n) of ${\text{H}}_{\text{2}}$ as follows:

$\begin{array}{c}\text{No}\text{.}\text{of}\text{moles=}\frac{\text{Given}\text{}\text{mass}}{\text{molar}\text{mass}}\\ =\frac{2.8}{2}\\ =1.4\end{array}$

The volume is converted from L to $m^3$ :

$\begin{array}{c}\text{Volume=10}\text{.0}\cancel{\text{L}}\times \left(\frac{1{\text{m}}^3}{1000\cancel{\text{L}}}\right)\\ =0.01{\text{m}}^3\end{array}$

The temperature is converted into Kelvin as follows:

$\begin{array}{c}\text{ 27}°\text{C=27 +273}\\ =300\text{K}\end{array}$

Put all values in equation (1) to calculate partial pressure of hydrogen gas.

$\begin{array}{l}{P}_{{\text{H}}_{\text{2}}}V=n_{{\text{H}}_{\text{2}}}RT\\ P_{{\text{H}}_{\text{2}}}=\frac{n_{{\text{H}}_{\text{2}}}RT}{V}\\ =\frac{1.4\cancel{\text{mol}}\left(8.31\text{J/}\cancel{\text{mol}}\text{}\cancel{\text{K}}\right)\left(300\cancel{\text{K}}\right)}{0.01{\text{m}}^3}\\ =\frac{3490.2\text{J}}{0.01{\text{m}}^3}\\ \left({\text{J/m}}^{\text{3}}=\text{1}\text{Pa}\right)\end{array}$

$=349020\text{Pa}$

Converting the values of pressure from Pa to mm Hg,

$\begin{array}{c}\text{Pressure=349020}\cancel{\text{Pa}}\left(\frac{760\text{mm}\text{Hg}}{102325\cancel{\text{Pa}}}\right)\\ =\frac{265255200\text{mm}\text{Hg}}{102325}\\ =2592.28\text{mm}\text{Hg}\end{array}$

If $3.2\text{g}$ of each gas is used then, pressure in the flask will be

$\begin{array}{c}{P}^{\ast }+P_{H_2}=26.7\text{mm}\text{Hg+2592}\text{.28}\text{mm}\text{Hg}\\ \text{=2618}\text{.98}\text{mm}\text{Hg}\end{array}$

The value of pressure is converted into atm as follows:

$\begin{array}{c}\text{Pressure=2618}\text{.98}\cancel{\text{mm}\text{}\text{Hg}}\left(\frac{1\text{atm}}{760\cancel{\text{mm}\text{}\text{Hg}}}\right)\\ =3.45\text{atm}\end{array}$

Therefore, pressure in the flask is $3.45\text{atm}$.