Question 4 (1 point) A balloon is placed in a closed vacuum chamber at STP. The chamber pressure is reduced by half. What happens to the balloon? ) It stays the same volume. ) It expands to four times the volume. It reduces in size by half. It doubles in volume. D View hint for Question 4

To answer this question, we need to apply **Boyle's Law**, which states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional. Mathematically:<br /><br />\[<br />P_1 V_1 = P_2 V_2<br />\]<br /><br />Where:<br />- \(P_1\) is the initial pressure,<br />- \(V_1\) is the initial volume,<br />- \(P_2\) is the final pressure,<br />- \(V_2\) is the final volume.<br /><br />### Given:<br />- The balloon starts at standard temperature and pressure (STP).<br />- The chamber pressure is reduced by half (\(P_2 = \frac{P_1}{2}\)).<br />- Temperature remains constant.<br /><br />### Step-by-step solution:<br />1. Rearrange Boyle's Law to solve for \(V_2\):<br /> \[<br /> V_2 = \frac{P_1 V_1}{P_2}<br /> \]<br /><br />2. Substitute \(P_2 = \frac{P_1}{2}\) into the equation:<br /> \[<br /> V_2 = \frac{P_1 V_1}{\frac{P_1}{2}} = V_1 \cdot 2<br /> \]<br /><br />3. This shows that the volume of the balloon doubles when the pressure is halved.<br /><br />### Final Answer:<br />**The balloon doubles in volume.**<br /><br />Correct option: **"It doubles in volume."**