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

The answer to your question is V2 = 3.7 L

Explanation:

Data

Volume 1 = 3.5 L

Temperature 1 = 25°C

Temperature 2 = 40°C

Pressure = cte.

Volume 2 = ?

Formula

To solve this problem use the Charles' law

             V1/T1 = V2/T2

- Solve for V2

             V2 = V1T2 / T1

- Convert temperature to °K

Temperature 1 = 25 + 273 = 298°K

Temperature 2 = 40 + 273 = 313°K

- Substitution

              V2 = (3.5)(313) / 298

- Simplification

              V2 = 1095.5 / 298

- Result

              V2 = 3.7 L

The new volume of the balloon is 3.68 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 3.50 L

The temperature = 25.0 °C = 298 K

The temperature changes to 40.0 °C = 313 K

The pressure remains constant at 1.0 atm

Step 2: Calculate the new volume of the balloon

V1/T1 = V2/T2

⇒with V1 = the initial volume of the balloon = 3.50 L

⇒with T1 = the initial temperature = 298 K

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the new temperature = 313 K

3.50 L / 298 K = V2 / 313 K

V2 = 3.68 L

The new volume of the balloon is 3.68 L