Q5.23P - 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) Half the gas escapes (at constant P and T).

(b) The initial pressure is 722 torr, and the final pressure is 0.950 atm; the initial temperature is 32^oF, and the final temperature is 273 K.

(c) Both the pressure and temperature decrease to one-fourth of their initial values.

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 is decreased by 0.99 times of its initial volume.

3. The volume of the gas remains constant.

Step 1: Subpart (a) Effect on volume if half the gas escapes (at constant P and T).

The ideal gas law equation is,

$\frac{\mathbf{P}\mathbf{V}}{\mathbf{n}\mathbf{T}}\mathbf{=}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathbf{ }\mathit{t}\mathit{a}\mathit{n}\mathbf{ }\mathit{t}$

Here,

P is pressure.

V is volume.

n is the number of moles.

T is temperature.

It is given that P and T are constant, then the ideal gas equation can be written as,

$$\begin{gathered} \frac{{\mathbf{PV}}_{\mathbf{1}}}{{\mathbf{n}}_{\mathbf{1}}\mathbf{T}}\mathbf{=}\frac{{\mathbf{PV}}_{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{2}}\mathbf{T}} \\ \frac{{\mathbf{v}}_{\mathbf{2}}}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{1}}} \end{gathered}$$

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

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

${\mathit{n}}_{\mathbf{1}}$ is the initial moles.

${\mathit{n}}_{\mathbf{2}}$ is the final moles.

The final volume of the gas is,

$$\begin{gathered} \frac{{\mathbf{v}}_{\mathbf{2}}}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\frac{{\displaystyle \frac{\mathbf{n}}{\mathbf{2}}}}{\mathbf{n}} \\ {\mathit{v}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{V}}_{\mathbf{1}} \end{gathered}$$

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

Step 2: Effect on volume if initial pressure is 722 torr, and the final pressure is 0.950 atm; the initial temperature is 32 ° F , and the final temperature is 273 K.

The ideal gas law equation is,

$\frac{\mathbf{PV}}{\mathbf{nT}}\mathbf{=}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$

The ideal gas equation for the given condition can be written as,

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

Here,

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

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

${\mathit{P}}_{\mathbf{1}}$ is the initial pressure (722 torr).

${\mathit{P}}_{\mathbf{2}}$ is the initial pressure (0.950 atm).

${\mathit{T}}_{\mathbf{1}}$ is the initial temperature ($\mathbf{32}\mathbf{°}\mathit{F}\mathbf{=}\mathbf{273}\mathbf{.}\mathbf{15}\mathit{K}$).

${\mathit{T}}_{\mathbf{2}}$ is the final temperature (273 K).

Also,

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

$$\begin{gathered} {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{\times }\frac{{\mathbf{T}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{atm}\mathbf{\times }{\mathbf{V}}_{\mathbf{1}}}{\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{K}}\mathbf{\times }\frac{\mathbf{273}\mathbf{K}}{\mathbf{0}\mathbf{.}\mathbf{950}\mathbf{atm}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}{\mathbf{99}}_{\mathbf{1}} \end{gathered}$$

Thus, the final volume of the gas is decreased by 0.99 times of its initial volume. 

Step 3: Effect on volume if both the pressure and temperature decrease to one-fourth of their initial values

The ideal gas law equation is,

$\frac{\mathbf{PV}}{\mathbf{nT}}\mathbf{=}\mathbf{constant}$

The ideal gas equation for the given condition can be written as,

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

The final volume of the gas is,

$\begin{array}{c}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{V}}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{P}\left(\frac{T}{4}\right)}{\mathbf{T}\left(\frac{P}{4}\right)}\\ {\mathbf{V}}_{\mathbf{2}}\mathbf{=}{\mathbf{V}}_{\mathbf{1}}\end{array}$

Thus, the final volume of the gas remains same as the initial volume.