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:
We are given the initial volume and temperature of a gas sample, and we need to find its new volume when the temperature decreases by a specific absolute amount, assuming constant pressure.

Step 2: Key Formula or Approach:
According to Charles's Law, the volume of a given mass of an ideal gas is directly proportional to its absolute temperature at constant pressure: $$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$ where temperatures must always be converted to the Kelvin scale ($T(K) = T(^\circ\text{C}) + 273$).

Step 3: Detailed Explanation:
Given values: Initial temperature, $T_1 = 0^\circ\text{C} = 0 + 273 = 273\ \text{K}$ Initial volume, $V_1 = 2\ \text{dm}^3$ The temperature decreases by $272^\circ\text{C}$, meaning the final temperature is: $T_2 = 0^\circ\text{C} - 272^\circ\text{C} = -272^\circ\text{C}$ Converting $T_2$ to Kelvin: $T_2 = -272 + 273 = 1\ \text{K}$ Now, substitute the values into the Charles's Law equation: $$\frac{2}{273} = \frac{V_2}{1}$$ $$V_2 = \frac{2}{273}\ \text{dm}^3$$

Step 4: Final Answer:
The final volume of the gas is $(2/273)\ \text{dm}^3$, matching option (D).