Chapter 12, Problem 1ICA

(a)

To determine

Find the slope units.

Answer to Problem 1ICA

The units of slope is $\underset{\_}{\text{L}/\text{K}}$.

Explanation of Solution

Calculation:

Refer to the graph in the respective question. The expression is,

$V=0.0175T$

Rearrange the expression.

$\frac{V}{T}=0.0175$

From the graph, the unit of volume (V) is L and temperature (T) is K. Therefore, the unit of slope (0.0175) is $\text{L}/\text{K}$.

Conclusion:

Thus, the units of slope is $\underset{\_}{\text{L}/\text{K}}$.

(b)

To determine

Find the mass of gas in terms of grams.

Answer to Problem 1ICA

The mass of nitrogen gas in terms of grams is $\underset{\_}{7.17\text{g}{\text{N}}_2}$.

Explanation of Solution

Given data:

Molecular weight (MW) of nitrogen is 28 grams per mole.

Pressure (P)is 1.2 atmospheres.

Formula used:

Consider the ideal gas law relationship.

$PV=nRT$ (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$n$ is number of moles.

$R$ is gas constant , which has standard value as $0.08206\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$

Calculation:

Refer to the graph in the respective question. The expression is,

$V=\left(0.0175\text{L}/\text{K}\right)T$

Modify equation (1) as follows.

$n=\frac{VP}{RT}$

Substitute $\left(0.0175\text{L}/\text{K}\right)T$ for $V$,

$\begin{array}{c}n=\frac{\left[\left(0.0175\text{L}/\text{K}\right)T\right]P}{RT}\\ =\frac{\left(0.0175\text{L}/\text{K}\right)P}{R}\end{array}$

Substitute 1.2 atm for P and $0.08206\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$ for $R$,

$\begin{array}{c}n=\frac{\left(0.0175\text{L}/\text{K}\right)\left(1.2\text{atm}\right)}{\left(0.08206\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}\right)}\\ =\frac{0.021}{0.08206}\text{mol}\\ =0.2559\text{mol}\\ \cong 0.256\text{mol}\end{array}$

Consider the general expression for mass in terms of molecular weight.

$m=n\left(\text{MW}\right)$ (2)

Substitute $28\text{g}/\text{mol}$ for MW and 0.256 mol for $n$ to obtain mass of the nitrogen gas.

$\begin{array}{c}m=\left(0.256\text{mol}\right)\left(28\text{g}{\text{N}}_2/\text{mol}\right)\\ =7.168\text{g}{\text{N}}_2\\ \cong 7.17\text{g}{\text{N}}_2\end{array}$

Conclusion:

Thus, the mass of nitrogen gas in terms of grams is $\underset{\_}{7.17\text{g}{\text{N}}_2}$.

(c)

To determine

Find the pressure of tank in terms of atmospheres.

Answer to Problem 1ICA

The pressure of tank in terms of atmospheres is 1.47 atm.

Explanation of Solution

Given data:

Molecular weight (MW) of oxygen is 32 grams per mole.

Mass of oxygen (m)is 10 grams.

Calculation:

Refer to the graph in the respective question.

Modify equation (2).

$n=\frac{m}{\text{MW}}$

Substitute 10 g for $m$ and $32\text{g}/\text{mol}$ for MW to obtain n.

$\begin{array}{c}n=\frac{10\text{g}}{32\text{g}/\text{mol}}\\ =0.3125\text{mol}\end{array}$

Modify equation (1).

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

Substitute $\left(0.0175\text{L}/\text{K}\right)T$ for $V$ and modify the expression as follows.

$\begin{array}{c}P=\frac{nRT}{\left[\left(0.0175\text{L}/\text{K}\right)T\right]}\\ =\frac{nR}{\left(0.0175\text{L}/\text{K}\right)}\end{array}$

Substitute 0.3125 mol for n and $0.08206\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$ for $R$ to obtain the value of pressure.

$\begin{array}{c}P=\frac{nR}{\left(0.0175\text{L}/\text{K}\right)}\\ =\frac{\left(0.3125\text{mol}\right)\left(0.08206\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}\right)}{\left(0.0175\text{L}/\text{K}\right)}\\ =\frac{0.0256}{0.0175}\text{atm}\\ =1.4628\text{atm}\end{array}$

$P\cong 1.47\text{atm}$

Conclusion:

Thus, the pressure of tank in terms of atmospheres is 1.47 atm.