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?

Short answer

Answer

The final volume of the sample is 56 L.

Step-by-step solution

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)

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.