Chapter 12, Problem 1ICA

The graph shows the ideal gas law relationship (PV =nRT) between volume (V) and temperature (T).

1. a. What are the units of the slope (0.0175)?
2. b. If the tank has a pressure of 2.4 atmospheres and is filled with nitrogen (formula, N2; molecular weight, 28 grams per mole), what is the mass of gas in the tank in units of grams?
3. c. If the tank is filled with 20 grams of oxygen (formula, o₂; molecular weight, 32 grams per mole), what is the pressure of the tank (P) in units of atmospheres?

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 $\underset{\_}{\text{L}/\text{K}}$.

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 14.4 g.

Explanation of Solution

Given data:

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

Pressure (P)is 2.4 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 $0.0821\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$

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 2.4 atm for P and $0.0821\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$ for $R$,

$\begin{array}{c}n=\frac{\left(0.0175\text{L}/\text{K}\right)\left(2.4\text{atm}\right)}{\left(0.0821\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}\right)}\\ =\frac{0.042}{0.0821}\text{mol}\\ =0.51157\text{mol}\\ \cong 0.512\text{mol}\end{array}$

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 0.512 mol for $n$,

$\begin{array}{c}m=\left(0.512\text{mol}\right)\left(28\text{g}/\text{mol}\right)\\ =14.336\text{g}\\ \cong 14.4\text{g}\end{array}$

Conclusion:

Thus, the mass of nitrogen gas in terms of grams is 14.4 g.

To determine

Find the pressure of tank in terms of atmospheres.

Answer to Problem 1ICA

The pressure of tank in terms of atmospheres is 3.0 atm.

Explanation of Solution

Given data:

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

Mass of oxygen (m)is 20 grams.

Calculation:

Refer to the graph in the respective question.

Modify equation (2).

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

Substitute 20 g for $m$ and $32\text{g}/\text{mol}$ for MW,

$\begin{array}{c}n=\frac{20\text{g}}{32\text{g}/\text{mol}}\\ =0.625\text{mol}\end{array}$

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 0.625 mol for n and $0.0821\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$ for $R$ to obtain the value of pressure.

$\begin{array}{c}P=\frac{nR}{\left(0.0175\text{L}/\text{K}\right)}\\ =\frac{\left(0.625\text{mol}\right)\left(0.0821\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}\right)}{\left(0.0175\text{L}/\text{K}\right)}\\ =\frac{0.0513}{0.0175}\text{atm}\\ =2.93\text{atm}\end{array}$

$P\cong 3.0\text{atm}$

Conclusion:

Thus, the pressure of tank in terms of atmospheres is 3.0 atm.