An expandable balloon contains 3.50 dm3 of helium at 101. kPa and 23.0oC. What volume will the balloon occupy when the pressure drops to 85.0 kPa and the temperature is 10.0oC?

3.98dm³

Explanation:

Using combined gas law equation:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (dm³)

V2 = final volume (dm³)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information in this question:

V1 = 3.5dm³

V2 = ?

P1 = 101 kPa

P2 = 85.0 kPa

T1 = 23.0°C = 23 + 273 = 296K

T2 = 10.0°C = 10 + 273 = 283K

Using P1V1/T1 = P2V2/T2

101 × 3.5/296 = 85 × V2/283

353.5/296 = 85V2/283

Cross multiply

296 × 85V2 = 353.5 × 283

25,160V2 = 100,040.5

V2 = 100,040.5 ÷ 25,160

V2 = 3.98dm³

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.