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

Hi there! :)

$\large\boxed{x = 2.49 L}$

Use the proportion for Charles' Law where:

$\frac{v_{1}}{t_{1}}= \frac{v_{2}}{t_{2}}$

v1 = initial volume

t1 = initial temperature

v2 = final volume

v2 = final temperature

Substitute in the given values into the proportion:

v1 = 4.39 L

t1 = 44° C

t2 = 25°C

v2 = x L

Set up the proportion:

$\frac{4.39}{44} = \frac{x}{25}$

Cross multiply:

$25 * 4.39 = 44x\\\\109.75 = 44x\\\\x = 2.49 L$

Answer 2

Answer:

The new volume will be approximately 2.49 L.

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.