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

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.