Answer to Question #345339 in General Chemistry for Grogs

Question

A 12.0 L balloon at 84 C is cooled until it becomes 5.0 L. To what temperature was the balloon cooled?

Answer 1

Given:

V₁ = 12.0 L

T₁ = 84°C + 273.15 = 357.15 K

P = constant

V₂ = 5.0 L

T₂ = unknown

Formula: V₁/T₁ = V₂/T₂

Solution:

Since the pressure and amount of gas remain unchanged, Charles's law can be used.

Charles's law can be expressed as: V₁ / T₁ = V₂ / T₂

V₁T₂ = V₂T₁

To find the resulting temperature, solve the equation for T₂:

T₂ = V₂T₁ / V₁

T₂ = (5.0 L × 357.15 K) / (12.0 L) = 148.81 K

T₂ = 148.81 K (or −124.34°C)

Answer: The balloon was cooled to −124.34°C