Question

A balloon contains 2.27 L air and has a pressure of $1.013 \times 10^5 \text{ Nm}^{-2}$. The balloon rises to a certain height and expands to volume of 4540 mL. What is the final pressure of air in balloon?

• A. $2.026 \times 10^2 \text{ Nm}^{-2}$
• B. $5.065 \times 10^4 \text{ Nm}^{-2}$
• C. $4.540 \times 10^4 \text{ Nm}^{-2}$
• D. $5.065 \times 10^{-4}$ $\text{ Nm}^{-2}$

Hint:

Boyle's Law states that pressure and volume share an inverse ratio. Since the volume doubles perfectly from $2.27\text{ L}$ to $4.54\text{ L}$ ($4540\text{ mL}$), the internal pressure must decrease by exactly half. Dividing $1.013 \times 10^5$ by 2 instantly gives $5.065 \times 10^4\text{ Nm}^{-2}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the initial volume ($V_1$) and initial pressure ($P_1$) of air trapped inside a balloon. As the balloon ascends, it expands to a known final volume ($V_2$). Assuming the temperature remains constant, we need to compute the new final pressure ($P_2$).

Step 2: Key Formula or Approach:
According to Boyle's Law, for a fixed mass of gas at a constant temperature, the pressure is inversely proportional to its volume: $$P_1 V_1 = P_2 V_2 \implies P_2 = \frac{P_1 V_1}{V_2}$$ Before substituting, ensure that both volume terms share identical dimensional units.

Step 3: Detailed Explanation:
Given variables: $P_1 = 1.013 \times 10^5 \text{ Nm}^{-2}$ $V_1 = 2.27 \text{ L}$ $V_2 = 4540 \text{ mL} = \frac{4540}{1000} \text{ L} = 4.54 \text{ L}$ Let's substitute these parameters into Boyle's Law: $$P_2 = \frac{(1.013 \times 10^5 \text{ Nm}^{-2}) \times 2.27 \text{ L}}{4.54 \text{ L}}$$ Notice the mathematical relationship between the volume terms: $4.54$ is exactly double $2.27$ ($\frac{4.54}{2.27} = 2$): $$P_2 = \frac{1.013 \times 10^5}{2}$$ $$P_2 = 0.5065 \times 10^5 = 5.065 \times 10^4 \text{ Nm}^{-2}$$

Step 4: Final Answer:
The final pressure of air inside the balloon is $5.065 \times 10^4 \text{ Nm}^{-2}$, which corresponds to option (B).