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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.