Question

A gas occupies $11.2\ \text{dm}^3$ at $105\ \text{kPa}$. What is its volume if pressure is increased to $210\ \text{kPa}$?

• A. $22.4\ \text{dm}^3$
• B. $33.6\ \text{dm}^3$
• C. $5.6\ \text{dm}^3$
• D. $16.8\ \text{dm}^3$

Hint:

Boyle's law tells us that pressure and volume are inversely proportional. Since the pressure was doubled ($105 \rightarrow 210$), the volume must be cut exactly in half! Half of $11.2$ is $5.6$—no extensive scratchpad math required.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the initial volume ($V_1$) and initial pressure ($P_1$) of a fixed mass of gas at constant temperature. We need to find the final volume ($V_2$) when the pressure is raised to a new value ($P_2$).

Step 2: Key Formula or Approach:
According to Boyle's law, for a given mass of an ideal gas at constant temperature, the volume is inversely proportional to its pressure. The product of pressure and volume remains constant: $$ P_1 V_1 = P_2 V_2 $$ Rearranging to calculate the final volume $V_2$: $$ V_2 = \frac{P_1 V_1}{P_2} $$

Step 3: Detailed Explanation:
Identify the values from the problem statement:

• Initial pressure, $P_1 = 105\ \text{kPa}$

• Initial volume, $V_1 = 11.2\ \text{dm}^3$

• Final pressure, $P_2 = 210\ \text{kPa}$
Substitute these values directly into the Boyle's law equation: $$ V_2 = \frac{105\ \text{kPa} \times 11.2\ \text{dm}^3}{210\ \text{kPa}} $$ Notice that $210$ is exactly twice $105$ ($\frac{105}{210} = \frac{1}{2}$): $$ V_2 = \frac{11.2}{2} = 5.6\ \text{dm}^3 $$

Step 4: Final Answer:
The volume of the gas when pressure increases to $210\ \text{kPa}$ is $5.6\ \text{dm}^3$, corresponding to option (C).