Chapter 7, Problem 7.53EP

(a)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

$\begin{array}{l}PV=nRT\\ \text{Where,}\\ P=\text{Pressure}\text{}\text{of}\text{gas}\\ V=\text{Volume}\text{}\text{of}\text{gas}\\ T=\text{Temperature}\text{of}\text{gas}\\ n=\text{Number}\text{of}\text{moles}\text{of}\text{gas}\\ R=\text{Ideal}\text{gas}\text{constant}\\ \end{array}$

Explanation of Solution

Record the given data,

$\begin{array}{l}\text{P = ?}\\ \text{V = 4}\text{.00L}\\ \text{n = 0}\text{.72 mole}\\ \text{T = 40}°\text{C converted to 313K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{\left(0.72\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(313\cancel{K}\right)}{\left(4.00\cancel{L}\right)}=\text{4}\text{.6 atm}$

Therefore, the new pressure is $\text{4}\text{.6 atm}.$

(b)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

$\begin{array}{l}PV=nRT\\ \text{Where,}\\ P=\text{Pressure}\text{}\text{of}\text{gas}\\ V=\text{Volume}\text{}\text{of}\text{gas}\\ T=\text{Temperature}\text{of}\text{gas}\\ n=\text{Number}\text{of}\text{moles}\text{of}\text{gas}\\ R=\text{Ideal}\text{gas}\text{constant}\\ \end{array}$

Explanation of Solution

Record the given data,

$\begin{array}{l}\text{P = ?}\\ \text{V = 4}\text{.00L}\\ \text{n = 4}\text{.5 mole}\\ \text{T = 40}°\text{C converted to 313K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{\left(4.5\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(313\cancel{K}\right)}{\left(4.00\cancel{L}\right)}=29\text{ atm}$

Therefore, the new pressure is $29\text{ atm}.$

(c)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

$\begin{array}{l}PV=nRT\\ \text{Where,}\\ P=\text{Pressure}\text{}\text{of}\text{gas}\\ V=\text{Volume}\text{}\text{of}\text{gas}\\ T=\text{Temperature}\text{of}\text{gas}\\ n=\text{Number}\text{of}\text{moles}\text{of}\text{gas}\\ R=\text{Ideal}\text{gas}\text{constant}\\ \end{array}$

Explanation of Solution

Record the given data,

$\begin{array}{l}\text{P = ?}\\ \text{V = 4}\text{.00L}\\ \text{n = 0}\text{.72 g}convertedto\left(\text{0}{\text{.72 g O}}_{\text{2}}\times \frac{\text{1}{\text{.00 mole O}}_{\text{2}}}{\text{32}{\text{.0 g O}}_{\text{2}}}\right)=0.031moles\\ \text{T = 40}°\text{C converted to 313K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{\left(0.0225\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(313\cancel{K}\right)}{\left(4.00\cancel{L}\right)}=0.14\text{ atm}$

Therefore, the new pressure is $\text{0}\text{.14 atm}.$

(d)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

$\begin{array}{l}PV=nRT\\ \text{Where,}\\ P=\text{Pressure}\text{}\text{of}\text{gas}\\ V=\text{Volume}\text{}\text{of}\text{gas}\\ T=\text{Temperature}\text{of}\text{gas}\\ n=\text{Number}\text{of}\text{moles}\text{of}\text{gas}\\ R=\text{Ideal}\text{gas}\text{constant}\\ \end{array}$

Explanation of Solution

Record the given data,

$\begin{array}{l}\text{P = ?}\\ \text{V = 4}\text{.00L}\\ \text{n = 4}\text{.5 g}convertedto\left(\text{4}{\text{.5 g O}}_{\text{2}}\times \frac{\text{1}{\text{.00 mole O}}_{\text{2}}}{\text{32}{\text{.0 g O}}_{\text{2}}}\right)=0.14moles\\ \text{T = 40}°\text{C converted to 313K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{\left(0.14\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(313\cancel{K}\right)}{\left(4.00\cancel{L}\right)}=0.90\text{ atm}$

Therefore, the new pressure is $\text{0}\text{.90 atm}.$