A helium-filled balloon escapes a child's hand at sea level and 20.1 ∘C. Part A When it reaches an altitude of 3600 m, where the temperature is 5.9 ∘C and the pressure is only 0.72 atm , how will its volume compare to that at sea level?

The volume of the helium-filled balloon would have increased by 1.46 times.

From the gas law:

                         P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

Hence; V2/V1 = P1T2/P2T1

In this case, P1 = pressure at sea level = 1 atm, P2 = 0.72 atm, T1 = 20.1 = 293.1, and T2 = 5.9 = 278.9

V2/V1 = 1 x 293.1/0.72 x 278.9

              = 1.46

Therefore, V2 = 1.46V1

The volume of the balloon will have increased by 1.46x.

More on the gas law can be found here: link

The volume at an altitude of 3600m will be 1.320 times the volume at sea level.

Explanation:

PV/T=CONSTANT

V2/V1=(P1/P2)×(T2/T1)

V2/V1 = (1 / 0.72) × (5.9+273.15 / 20.1+273.15)

          = (1 / 0.72) × (279.05 / 293.25)

          = 1.389 × 0.951

V2 / V1 = 1.320

V2 = 1.320 V1

V1 = volume at sea level

V2 = volume at 3600m

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error