Question

A gas sample occupies 4.39 L at 44 ºC. What will be the new volume, using Charles’ Law, if the temperature decreases to 25 ºC?​

Answer 1

Answer:

Explanation:

We are given that a gas sample occupies 4.39 L at 44° C.

We are also given that we are going decrease the temperature to 25° C.

Charles' Law has a formula which shows the relationship between the volumes of a samples and the temperatures of the samples.

The proportion is shown as:

$\displaystyle \bullet \ \ \ \frac{V_1}{T_1} = \frac{V_2}{T_2},$

where V₁ and V₂ are the initial and final volume respectively and T₁ and T₂ are the initial and final temperature respectively.

This proportion can be used to determine what our unknown is. We need to know the final volume.

$\displaystyle \frac{4.39 \ \text{L}}{44 \ \text{C}} = \frac{\text{x}}{25 \ \text{C}}\\\\\\\frac{4.39}{44} = \frac{x}{25}\\\\\\4.39 \times 25 = 44 \times x\\\\\\44x = 109.75\\\\\\\frac{44x}{44}=\frac{109.75}{44}\\\\\\x = 2.49432 \approx \boxed{2.49 \ \text{L}}$

Therefore, if we reduce the temperature to 25 °C, we will be reducing the volume of the substance to 2.49 L.

Answer 2

Answer:

$\begin{gathered}\begin{gathered}\implies \displaystyle \dfrac{4.39 \ \text{L}}{44 \ \text{C}} = \dfrac{\text{x}}{25 \ \text{C}}\\\\\\\ \implies\dfrac{4.39}{44} = \dfrac{x}{25}\\\\\\ \implies 4.39 \times 25 = 44 \times x\\\\\\ \implies 44x = 109.75\\\\\\ \implies \dfrac{44x}{44}=\dfrac{109.75}{44}\\\\\\x = 2.49432 \approx \boxed{2.49 \ \text{L}}\end{gathered}\end{gathered}$