Question

A balloon that had a volume of 3. 50 L at 25. 0°C is placed in a hot room at 40. 0°C. If the pressure remains constant at 1. 00 atm, what is the new volume of the balloon in the hot room? Use mc005-1. Jpg. 2. 19 L 3. 33 L 3. 68 L 5. 60 L.

Answer 1

According to Charles's ideal law of gas volume of occupied gas is directly proportionate to the temperature where the pressure remains unaffected.

The volume of the balloon will be:

Option 3. 3.68 L

The volume can be estimated by:

According to Charles's law :

$\rm \dfrac {V_{1}}{T_{1}}= \dfrac{V_{2}}{T_{2}}$

Where,

• V₁ = 3.5 L

• $\begin{aligned}\rm T_{1} &= 25 + 273 \\&= 298 \;\rm K \end{aligned}$

• V₂ = ?

• $\begin{aligned}\rm T_{2} &= 40+273 \\&= 313 \rm \;K \end{aligned}$

• P = 1.00 atm

Putting values in the equation:

$\rm \dfrac{3.5 \;L }{298 \;K} = \dfrac{V_{2} }{313 \;K}$

Solving further for V₂ :

$\begin{aligned}\rm V_{2} &= \rm (3.5 \;L)\times \dfrac {313 \;K }{298 \;K} \\\\\rm V_{2} &= 3.68 \;\rm L\end{aligned}$

Therefore the volume of the balloon is 3.68 L.

To learn more about Charles's law of gas equation follow the link:

brainly.com/question/10511054