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

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

: )