Q5.25P - Chemistry: Molecular Nature Of Matter And Change

Question

A 93-L sample of dry air cools from $\mathbf{145}\mathbf{°}\mathit{C}$ to $\mathbf{-}\mathbf{22}\mathbf{°}\mathit{C}$ while the pressure is maintained at 2.85 atm. What is the final volume?

Step-by-Step Solution

Answer

Answer

The final volume of the sample is 56 L.

Step 1: Ideal gas law

The ideal gas law equation is,

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

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

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

(Moles of the sample are fixed)

Step 2: Determination of final volume

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

Here,

${\mathit{V}}_{\mathbf{1}}$ is the initial volume (93 L).

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

${\mathit{T}}_{\mathbf{1}}$ is the initial temperature ($\mathbf{145}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{418}\mathbf{ }\mathit{K}$).

${\mathit{T}}_{\mathbf{2}}$ is the final temperature ($\mathbf{-}\mathbf{22}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{251}\mathbf{ }\mathit{K}$).

$$\begin{gathered} {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{\times }{\mathit{T}}_{\mathbf{2}} \\ {\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{93}\mathbf{ }\mathbf{L}}{\mathbf{418}\mathbf{ }\mathbf{K}}\mathbf{\times }\mathbf{251}\mathbf{ }\mathit{K} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{56}\mathbf{ }\mathit{L} \end{gathered}$$

Thus, the final volume is 56 L.