Q5.21P - Chemistry: Molecular Nature Of Matter And Change

Question

 What is the effect of the following on the volume of 1 mol of an ideal gas?

(a) The pressure is reduced by a factor of 4 (at constant T).

(b) The pressure changes from 760 torr to 202 kPa, and the temperature changes from $\mathbf{37}\mathbf{°}\mathit{C}$ to 155 K.

(c) The temperature changes from 305 K to $\mathbf{32}\mathbf{°}\mathit{C}$, and the pressure changes from 2 atm to 101 kPa.

Step-by-Step Solution

Answer

Answer

1. At constant temperature, if the pressure is reduced by a factor of 4, then the new volume of the will be 4 times its initial volume.

2. The new volume of the gas will be 1/4 of its initial volume.

3. The new volume of the gas will be double its initial volume.

Step 1: Subpart (a) Effect on the volume of 1 mol of an ideal gas when the pressure is reduced by a factor of 4.

The ideal gas law equation is,

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

At constant temperature, the pressure is reduced by the factor of 4, then the new ideal gas law equation is,

$\frac{\mathbf{P}}{\mathbf{4}}\mathit{V}\mathbf{\text{'}}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

Divide the above equation by the ideal gas law equation,

$$\begin{gathered} \frac{{\displaystyle \frac{\mathbf{P}}{\mathbf{4}}\mathbf{V}\mathbf{\text{'}}}}{\mathbf{PV}}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{nRT}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{V}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathit{V} \end{gathered}$$

Thus, at a constant temperature, if the pressure is reduced by a factor of 4, then the new volume will increase 4 times its initial volume.

Step 2: Subpart (b) Effect on the volume of 1 mol of an ideal gas when the pressure changes from 760 torr to 202 kPa, and the temperature change from 37 ° C to 155 K.

The ideal gas law equation is,

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

The ideal gas law equation for the given condition is,

$\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}{\mathbf{n}}_{\mathbf{2}}}$

Here,

${\mathit{n}}_{\mathbf{1}}$ and ${\mathit{n}}_{\mathbf{2}}$ is 1 mol.

The initial pressure ${\mathit{P}}_{\mathbf{1}}$ is 760 torr.

The final pressure ${\mathit{P}}_{\mathbf{2}}$ is 202 kPa.

The initial temperature ${\mathit{T}}_{\mathbf{1}}$ is $\mathbf{37}\mathbf{°}\mathit{C}$.

The final temperature ${\mathit{T}}_{\mathbf{2}}$ is 155K.

The initial volume is ${\mathit{V}}_{\mathbf{1}}$.

The final volume is ${\mathit{V}}_{\mathbf{2}}$.

Conversion of units,

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathit{k}\mathit{P}\mathit{a}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{ }\mathit{t}\mathit{o}\mathit{r}\mathit{r} \\ \mathbf{760}\mathbf{ }\mathit{t}\mathit{o}\mathit{r}\mathit{r}\mathbf{\times }\frac{\mathbf{1}\mathbf{kPa}}{\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{ }\mathbf{torr}}\mathbf{=}\mathbf{101}\mathbf{ }\mathit{k}\mathit{P}\mathit{a} \end{gathered}$$

Initial temperature, $\mathbf{37}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{310}\mathit{K}$

Now, the final volume of the gas is,

$$\begin{gathered} \frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}{\mathbf{n}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{1}}}\mathbf{\times }\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{101}\mathbf{kPa}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}}}{\mathbf{310}\mathbf{\times }\mathbf{1}\mathbf{mol}}\mathbf{\times }\frac{\mathbf{155}\mathbf{K}\mathbf{\times }\mathbf{1}\mathbf{mol}}{\mathbf{202}\mathbf{ }\mathbf{kPa}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}{\mathit{V}}_{\mathbf{1}} \end{gathered}$$

Thus, the new volume of the gas will be 1/4 to its initial volume.

Step 3: Subpart (c) Effect on volume of 1 mol of an ideal gas when temperature changes from 305 K to 32 ° C , and the pressure change from 2 atm to 101 kPa.

The ideal gas law equation for the given condition is,

$\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}{\mathbf{n}}_{\mathbf{2}}}$

Here,

${\mathit{n}}_{\mathbf{1}}$ and ${\mathit{n}}_{\mathbf{2}}$ is 1 mol.

The initial pressure ${\mathit{P}}_{\mathbf{1}}$ is 2 atm.

The final pressure ${\mathit{P}}_{\mathbf{2}}$ is 101 kPa.

The initial temperature ${\mathit{T}}_{\mathbf{1}}$ is 305 K.

The final temperature ${\mathit{T}}_{\mathbf{2}}$ is $\mathbf{32}\mathbf{°}\mathit{C}$.

The initial volume is ${\mathit{V}}_{\mathbf{1}}$.

The final volume is ${\mathit{V}}_{\mathbf{2}}$.

Conversion of units,

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathit{k}\mathit{P}\mathit{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00987}\mathbf{ }\mathit{a}\mathit{t}\mathit{m} \\ \mathbf{101}\mathit{k}\mathit{P}\mathit{a}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{00987}\mathbf{ }\mathbf{atm}}{\mathbf{1}\mathbf{kPa}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{a}\mathit{t}\mathit{m} \end{gathered}$$

Final temperature is $\mathbf{32}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{305}\mathit{K}$.

Now, the final volume of the gas is,

$$\begin{gathered} \frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}{\mathbf{n}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{1}}}\mathbf{\times }\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}{\mathbf{n}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}\mathbf{ }\mathbf{atm}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}}}{\mathbf{305}\mathbf{K}\mathbf{\times }\mathbf{1}\mathbf{mol}}\mathbf{\times }\frac{\mathbf{305}\mathbf{K}\mathbf{\times }\mathbf{1}\mathbf{mol}}{\mathbf{1}\mathbf{atm}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{2}{\mathit{V}}_{\mathbf{1}} \end{gathered}$$

Thus, the new volume of the gas will be double to its initial volume.