Question

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?

Answer 1

The new volume is 31.97 L

Step-by-step 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

Answer 2

The answer to your question is V2 = 31.97 L

Step-by-step 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