Q5.22P - 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) Temperature decreases from 800 K to 400 K (at constant P).

(b) Temperature increases from $\mathbf{250}\mathbf{°}\mathit{C}$ to $\mathbf{500}\mathbf{°}\mathit{C}$ (at constant P).

(c) Pressure increases from 2 atm to 6 atm (at constant T).

Step-by-Step Solution

Answer

Answer

1. The final volume of the gas becomes half of its initial volume.

2. The final volume of the gas becomes 1.48 times its initial volume.

3. The final volume of the gas becomes 1/3 of its initial volume.

Step 1: Subpart (a) Effect on volume if temperature decreases from 800 K to 400 K (at constant P).

According to Charles’ law, at constant pressure, the volume of the gas is directly proportional to the temperature of the gas.

$$\begin{gathered} \mathit{V}\mathbf{\infty }\mathit{T} \\ \frac{\mathbf{V}}{\mathbf{T}}\mathbf{=}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathbf{ }\mathit{t}\mathit{a}\mathit{n}\mathbf{ }\mathit{t}\left(P,n fised\right) \end{gathered}$$

The above equation can be written as,

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

Here,

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

${\mathit{V}}_{\mathbf{1}}$ is the final volume.

${\mathit{T}}_{\mathbf{1}}$ is the initial temperature (800 K).

 ${\mathit{T}}_{\mathbf{2}}$ is the final temperature (400 K).

The final volume of the gas is,

$$\begin{gathered} \frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{\times }{\mathit{T}}_{\mathbf{2}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{\mathbf{800}\mathbf{K}}\mathbf{\times }\mathbf{400}\mathit{K} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{V}}_{\mathbf{1}} \end{gathered}$$

Thus, the final volume of the gas becomes half of its initial volume.

Step 2: (b) Effect on volume if temperature increases from 250 ° C to 500 ° C (at constant P).

According to Charles’ law, at constant pressure, the volume of the gas is directly proportional to the temperature of the gas. It can also be written as

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

Here,

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

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

${\mathit{T}}_{\mathbf{1}}$ is the initial temperature ($\mathbf{250}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{523}\mathit{K}$).

${\mathit{T}}_{\mathbf{2}}$ is the final temperature ($\mathbf{500}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{773}\mathit{K}$).

The final volume of the gas is,

$$\begin{gathered} \frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{\times }{\mathit{T}}_{\mathbf{2}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{\mathbf{523}\mathbf{K}}\mathbf{\times }\mathbf{773}\mathit{K} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{48}{\mathit{V}}_{\mathbf{1}} \end{gathered}$$

Thus, the final volume of the gas becomes 1.48 times of its initial volume.

Step 3: Subpart (c) Effect on volume if pressure increases from 2 atm to 6 atm (at constant T).

According to Boyle’s law, at constant temperature, the volume of the gas is inversely proportional to the pressure of the gas.

$$\begin{gathered} \mathit{V}\mathbf{\infty }\frac{\mathbf{1}}{\mathbf{P}} \\ \mathit{P}\mathit{V}\mathbf{=}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathbf{ }\mathit{t}\mathit{a}\mathit{n}\mathbf{ }\mathit{t}\mathbf{(}\mathbf{T}\mathbf{,}\mathbf{n}\mathbf{ }\mathbf{fised}\mathbf{)} \end{gathered}$$

The above equation can be written as,

${\mathit{P}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}\mathbf{=}{\mathit{P}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}$

Here,

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

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

${\mathit{T}}_{\mathbf{1}}$ is the initial pressure ( 2 atm).

${\mathit{T}}_{\mathbf{2}}$ is the final pressure ( 6 atm).

The final volume of the gas is,

$$\begin{gathered} {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{P}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{\left(}\mathbf{2}\mathbf{atm}\mathbf{\right)}{\mathbf{V}}_{\mathbf{1}}}{\mathbf{6}\mathbf{atm}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathit{V}}_{\mathbf{1}} \end{gathered}$$

Thus, the final volume of the gas becomes 1/3 of its initial volume.