Q5.24P - Chemistry: Molecular Nature Of Matter And Change

Question

A sample of sulfur hexafluoride gas occupies 9.10 L at $\mathbf{198}\mathbf{°}\mathit{C}$. Assuming that the pressure remains constant, what temperature (in $\mathbf{°}\mathit{C}$) is needed to reduce the volume to 2.50 L?

Step-by-Step Solution

Answer

Answer

At constant pressure, the temperature is needed to reduce to $\mathbf{-}\mathbf{144}\mathbf{°}\mathit{C}$.

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 at constant pressure 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 temperature

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

Here,

${\mathit{V}}_{\mathbf{1}}$ is the initial volume (9.10 L).

${\mathit{V}}_{\mathbf{2}}$ is the final volume (2.50 L).

${\mathit{T}}_{\mathbf{1}}$ is the initial temperature ($\mathbf{198}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{471}\mathit{K}$).

${\mathit{T}}_{\mathbf{2}}$ is the final temperature.

$$\begin{gathered} {\mathit{T}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{\times }{\mathit{V}}_{\mathbf{2}} \\ {\mathit{T}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{471}\mathbf{ }\mathbf{K}}{\mathbf{9}\mathbf{.}\mathbf{10}\mathbf{ }\mathbf{L}}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{ }\mathit{L} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{129}\mathit{K} \\ {\mathit{T}}_{\mathbf{2}}\mathbf{=}\mathbf{129}\mathbf{-}\mathbf{273}\mathbf{=}\mathbf{-}\mathbf{144}\mathbf{°}\mathit{C} \end{gathered}$$

Thus, at constant pressure, the temperature is needed to reduce to $\mathbf{-}\mathbf{144}\mathbf{°}\mathit{C}$.