Question

If $\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$of argon occupies a 75.0-mL container at $\mathbf{26}\mathbf{.}\mathbf{°}\mathit{C}$, what is the pressure (in torr)?

Short answer

Answer

The pressure of the gas is 365 torr.

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,

$\mathit{P}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{V}}$

Here,

V is the volume ($\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{075}\mathbf{\hspace{0.17em}}\mathit{L}$).

T is the temperature ($\mathbf{26}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{299}\mathbf{}\mathit{K}$).

n is the number of moles ($\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}$).

Determination of the pressure.

The pressure of the gas is,

$$\begin{gathered} \mathit{P}\mathbf{=}\frac{\mathbf{n}\mathbf{R}\mathbf{T}}{\mathbf{V}} \\ \mathit{P}\mathbf{=}\frac{\left(1.47\times {10}^{-3}mol\right)\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0821}\mathbf{L}\mathbf{a}\mathbf{t}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}\mathbf{m}\mathbf{o}{\mathbf{l}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{299}\mathbf{}\mathbf{K}}{\mathbf{0}\mathbf{.}\mathbf{075}\mathbf{}\mathbf{L}} \\ \mathit{P}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{48}\mathbf{}\mathit{a}\mathit{t}\mathit{m} \end{gathered}$$

Conversion of the unit of pressure from atm to torr,

$\mathbf{1}\mathbf{}\mathit{t}\mathit{o}\mathit{r}\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00132}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$

And,

$\mathbf{0}\mathbf{.}\mathbf{481}\mathbf{}\mathit{a}\mathit{t}\mathit{m}\mathbf{\times }\frac{\mathbf{1}\mathbf{}\mathbf{torr}}{\mathbf{0}\mathbf{.}\mathbf{00132}}$

Thus, the pressure of the gas is 365 torr.