A balloon filled with helium has a volume of 30.0 L at a pressure of 100 kPa and a temperature at 15.0 Celsius. What will The volume of the balloon be if the temperature is increased at 80.0 Celsius and the pressure changes to 115 kPa?

The answer to your question is V2 = 31.97 L

Explanation:

Data

Volume 1 = 30 L

Pressure 1 = 100 kPa

Temperature 1 = 15°C

Volume 2 = ?

Pressure 2 = 115 kPa

Temperature 2 = 80°C

Process

1.- Convert temperature to °K

Temperature 1 = 15 + 273 = 288°K

Temperature 2 = 80 + 273 = 353°K

2.- Use the Combined gas law to solve this problem

                  P1V1/T1 = P2V2/T2

-Solve for V2

                 V2 = P1V1T2 / T1P2

-Substitution

                 V2 = (100 x 30 x 353) / (288 x 115)

- Simplification

                 V2 = 1059000 / 33120

- Result

                 V2 = 31.97 L

The new volume is 31.97 L

Explanation:

Step 1: Data given

The initial volume = 30.0 L

The initial pressure = 100 kPa

Temperature = 15.0 °C = 288 K

The temperature is increased to 80.0 °C = 353 K

The pressure increases to 115 kPa

Step 2: Calculate the new volume

P1*V1 /T1  = P2*V2/T2

⇒with P1 = the initial pressure = 100 kPa

⇒with V1 = the initial volume = 30.0 L

⇒with T1 = the initial temperature = 288 K

⇒with P2 = the increased pressure = 115 kPa

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the increased temperature = 353 K

(100 * 30.0)/288 = (115 * V2)/353

10.417 = (115 * V2)/353

V2 = 31.97 L

The new volume is 31.97 L

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