Answer to Question #113357 in General Chemistry for Stephanie

Question

A balloon measuring 3.51 L at 55 degrees Celsius is cooled to 5 degrees Celsius what is the new volume in L?

Answer 1

P₁ = P₂ = constant;

V₁ = 3.51 L;

T₁ = 55 °C = 328.15 K;

T₂ = 5 °C = 278.15 K;

V₂ = unknown.

Solution:

(P₁×V₁)/(T₁) = (P₂×V₂)/(T₂);

P₁ = P₂ = constant;

(V₁)/(T₁) = (V₂)/(T₂);

V₂ = (V₁×T₂)/(T₁);

V₂ = (3.51 L × 278.15 K)/(328.15 K) = 2.97 L;

Answer: 2.97 L