Chapter 7, Problem 7.54EP

(a)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for nitrogen 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,

$\begin{array}{l}\text{P = ?}\\ \text{V = 2}\text{.00L}\\ \text{n = 0}\text{.50 mole}\\ \text{T = 50}°\text{C converted to 323K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

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

$PV=nRT$

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

$P=\frac{\left(0.50\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(323\cancel{K}\right)}{\left(2.00\cancel{L}\right)}=6.6\text{ atm}$

Therefore, the new pressure is $6.6\text{ atm}.$

(b)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for nitrogen 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,

$\begin{array}{l}\text{P = ?}\\ \text{V = 2}\text{.00L}\\ \text{n = 2}\text{.3 mole}\\ \text{T = 50}°\text{C converted to 323K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

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

$PV=nRT$

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

$P=\frac{\left(2.3\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(323\cancel{K}\right)}{\left(2.00\cancel{L}\right)}=30\text{ atm}$

Therefore, the new pressure is $\text{30 atm}.$

(c)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for nitrogen 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,

$\begin{array}{l}\text{P = ?}\\ \text{V = 2}\text{.00L}\\ \text{n = 0}\text{.35 g}convertedto\left(\text{0}{\text{.35 g N}}_{\text{2}}\times \frac{\text{1}{\text{.00 mole N}}_{\text{2}}}{28{\text{ g N}}_{\text{2}}}\right)=0.0125moles\\ \text{T = 40}°\text{C converted to 323K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

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

$PV=nRT$

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

$P=\frac{\left(0.0125\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(323\cancel{K}\right)}{\left(2.00\cancel{L}\right)}=0.17\text{ atm}$

Therefore, the new pressure is $\text{0}\text{.17 atm}.$

(d)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for nitrogen 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,

$\begin{array}{l}\text{P = ?}\\ \text{V = 2}\text{.00L}\\ \text{n = 3}\text{.7 g}convertedto\left(\text{3}{\text{.7 g N}}_{\text{2}}\times \frac{\text{1}{\text{.00 mole N}}_{\text{2}}}{28{\text{ g N}}_{\text{2}}}\right)=0.0132moles\\ \text{T = 40}°\text{C converted to 323K}\\ \text{R}\text{=}\text{0}\text{.821 }atm.L/mole.K\end{array}$

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

$PV=nRT$

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

$P=\frac{\left(0.132\cancel{mole}\right)\left(0.0821atm.\cancel{L}/\cancel{mole.K}\right)\left(323\cancel{K}\right)}{\left(2.00\cancel{L}\right)}=1.8\text{ atm}$

Therefore, the new pressure is $1.8atm.$