Question

The volume of a gas at $0^\circ \text{C}$ is $2\ \text{dm}^3$. What is its volume if temperature is decreased by $272^\circ \text{C}$?

• A. $(3/272)\ \text{dm}^3$
• B. $(2/272)\ \text{dm}^3$
• C. $(4/273)\ \text{dm}^3$
• D. $(2/273)\ \text{dm}^3$

Hint:

Always double check the wording: "decreased by $272^\circ\text{C}$" means you subtract 272 from the initial value. Never perform gas law calculations using Celsius values directly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The task is to calculate the new volume of a gas when its temperature is lowered by a defined interval, while maintaining constant pressure.

Step 2: Key Formula or Approach:
Charles's Law governs this relationship: V₁/T₁ = V₂/T₂, where the temperatures must be expressed in Kelvin (K).

Step 3: Detailed Explanation:
Initial temperature T₁ = 0°C = 273 K, and initial volume V₁ = 2 dm³. The temperature drops by 272°C, so final T₂ = 0 - 272 = -272°C. Converting to Kelvin gives T₂ = -272 + 273 = 1 K. Applying Charles's Law: 2/273 = V₂/1, which solves to V₂ = 2/273 dm³.

Step 4: Final Answer:
The volume of the gas becomes (2/273) dm³, corresponding to option (D).