Question

a scuba diver with a lung capacity of 5.2 l inhales at a depth of 41 m and a pressure of 5.1 atm. part a if the diver were to ascend to the surface (where the pressure is 1.0 atm) while holding her breath, to what volume would the air in her lungs expand? (assume constant temperature.) express your answer in liters to two significant figures. v = \boxed{\space} l

Explanation:

Step1: Recall Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, the product of pressure and volume is constant, i.e., \( P_1V_1 = P_2V_2 \).
Here, \( P_1 = 5.1 \, \text{atm} \), \( V_1 = 5.2 \, \text{L} \), and \( P_2 = 1.0 \, \text{atm} \). We need to find \( V_2 \).

Step2: Rearrange the formula to solve for \( V_2 \)

From \( P_1V_1 = P_2V_2 \), we can solve for \( V_2 \) by dividing both sides by \( P_2 \):

$$ V_2 = \frac{P_1V_1}{P_2} $$

Step3: Substitute the given values

Substitute \( P_1 = 5.1 \, \text{atm} \), \( V_1 = 5.2 \, \text{L} \), and \( P_2 = 1.0 \, \text{atm} \) into the formula:

$$ V_2 = \frac{5.1 \, \text{atm} \times 5.2 \, \text{L}}{1.0 \, \text{atm}} $$

Step4: Calculate the value

First, multiply \( 5.1 \) and \( 5.2 \): \( 5.1 \times 5.2 = 26.52 \). Then divide by \( 1.0 \): \( V_2 = 26.52 \, \text{L} \). Now, round to two significant figures. The number \( 26.52 \) rounded to two significant figures is \( 27 \) (since the third digit is \( 5 \), we round up the second digit). Wait, actually, let's check the significant figures. The given values: \( P_1 = 5.1 \) (two sig figs), \( V_1 = 5.2 \) (two sig figs), \( P_2 = 1.0 \) (two sig figs). So the result should have two sig figs. Wait, \( 5.1 \times 5.2 = 26.52 \), divided by \( 1.0 \) is \( 26.52 \). Rounding to two significant figures: the first two significant figures are \( 2 \) and \( 6 \), the next digit is \( 5 \), so we round up the \( 6 \) to \( 7 \), giving \( 27 \). Wait, but let's do the calculation more accurately. \( 5.1 \times 5.2 = (5 + 0.1)(5 + 0.2) = 25 + 1 + 0.5 + 0.02 = 26.52 \). Then \( 26.52 / 1.0 = 26.52 \). Rounded to two significant figures: \( 27 \) (because the third digit is \( 5 \), so we round the second digit \( 6 \) up to \( 7 \)). Wait, but maybe I made a mistake. Wait, \( 5.1 \) and \( 5.2 \) are two significant figures each, so the product should have two significant figures? Wait, no: when multiplying, the number of significant figures in the result is equal to the number of significant figures in the least precise measurement. Both \( 5.1 \) and \( 5.2 \) have two significant figures, so the product has two significant figures? Wait, no, actually, \( 5.1 \times 5.2 = 26.52 \), which is more precise, but when we consider significant figures, we look at the original data. Wait, maybe the problem expects us to keep two significant figures. So \( 26.52 \) rounded to two significant figures is \( 27 \) (since the first two are \( 2 \) and \( 6 \), the next digit is \( 5 \), so round up \( 6 \) to \( 7 \)). Wait, but let's check again. Wait, \( 5.1 \) is two sig figs, \( 5.2 \) is two sig figs, so the result should be two sig figs. So \( 26.52 \) rounded to two sig figs is \( 27 \) (because \( 26.52 \) is closer to \( 27 \) than \( 26 \) when rounding to two sig figs? Wait, no. Wait, two significant figures: the first digit is \( 2 \), the second is \( 6 \), the third is \( 5 \). So when rounding to two sig figs, we look at the third digit. If it's \( 5 \) or more, we round up the second digit. So \( 26.52 \) rounded to two sig figs is \( 27 \) (since \( 6 + 1 = 7 \), and the first digit remains \( 2 \)). So \( V_2 = 27 \, \text{L} \) (to two significant figures). Wait, but let's do the calculation again. \( 5.1 \times 5.2 = 26.52 \), divided by \( 1.0 \) is \( 26.52 \). Rounded to two significant figures: \( 27 \) (because the third digit is \( 5 \), so we round up the second digit from \( 6 \) to \( 7 \)).

Answer:

\( 27 \)