Answer to Question #349309 in General Chemistry for JAYDEN

Question

A child has a toy balloon with a volume of 2.80 liters. The

temperature of the balloon when it was filled was 25° C and the

pressure was 2.00 atm. If the child were to let go of the balloon

and it rose 3 kilometers into the sky where the pressure is

0.578 atm and the temperature is -9° C, what would the new

volume of the balloon be?

Answer 1

Given:

P₁ = 2.00 atm

V₁ = 2.80 L

T₁ = 25°C + 273.15 = 298.15 K

P₂ = 0.578 atm

V₂ = unknown

T₂ = −9°C + 273.15 = 264.15 K

Formula: P₁V₁/T₁ = P₂V₂/T₂

Solution:

Combined gas law can be used.

Combined gas law can be expressed as:

P₁V₁/T₁ = P₂V₂/T₂

Cross-multiply to clear the fractions:

P₁V₁T₂ = P₂V₂T₁

Divide to isolate V₂:

V₂ = (P₁V₁T₂) / (P₂T₁)

Plug in the numbers and solve for V₂:

V₂ = (2.00 atm × 2.80 L × 264.15 K) / (0.578 atm × 298.15 K) = 8.58 L

V₂ = 8.58 L

Answer: The new volume of the balloon would be 8.58 liters