Question

a sealed syringe contains neon gas (ne) at a pressure of 5.51 atmospheres (atm). the plunger of the syringe is pulled back until the pressure and volume of the gas in the syringe are 4.45 atmospheres and 0.046 liters, respectively. what was the initial volume of the ne gas in the syringe before the plunger was pulled back? assume ideal gas behavior and a constant temperature. write your answer to the correct number of significant figures. round if necessary. liters save answer

Explanation:

Step1: Recall Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume.

Step2: Identify known values

We know that \( P_1 = 5.51 \, \text{atm} \), \( P_2 = 4.45 \, \text{atm} \), and \( V_2 = 0.046 \, \text{L} \). We need to find \( V_1 \).

Step3: Rearrange Boyle's Law to solve for \( V_1 \)

From \( P_1V_1 = P_2V_2 \), we can rearrange to get \( V_1=\frac{P_2V_2}{P_1} \).

Step4: Substitute the known values into the formula

Substitute \( P_2 = 4.45 \, \text{atm} \), \( V_2 = 0.046 \, \text{L} \), and \( P_1 = 5.51 \, \text{atm} \) into the formula:
\( V_1=\frac{4.45\times0.046}{5.51} \)
First, calculate the numerator: \( 4.45\times0.046 = 0.2047 \)
Then, divide by the denominator: \( V_1=\frac{0.2047}{5.51}\approx0.037 \, \text{L} \) (rounded to three significant figures, since the given values have three significant figures: 5.51 (three), 4.45 (three), 0.046 (two? Wait, 0.046 has two significant figures? Wait, 0.046: the leading zeros are not significant, so 4 and 6 are significant, so two. But 5.51 and 4.45 have three. When multiplying/dividing, the result should have the same number of significant figures as the least precise measurement. Here, 0.046 has two, 4.45 has three, 5.51 has three. Wait, no: 4.45×0.046: 4.45 has three, 0.046 has two, so the product has two? Wait, no, 0.046 is two significant figures? Wait, 0.046: the first non - zero digit is 4, then 6, so two significant figures. 4.45 has three. So 4.45×0.046: 3×2 = 2 significant figures in the product? Wait, no, the rule is that when multiplying or dividing, the result should have the same number of significant figures as the number with the least number of significant figures. So 4.45 (three) ×0.046 (two) = 0.2047, which should be rounded to two significant figures? Wait, no, maybe I made a mistake. Wait, 0.046 liters: is that two or three? Wait, 0.046: the 4 and 6 are significant, so two. 5.51 is three, 4.45 is three. So when we do \( V_1=\frac{4.45\times0.046}{5.51} \), the numerator is 4.45×0.046. 4.45 has three, 0.046 has two, so the numerator should be rounded to two significant figures? Wait, no, actually, we should keep more digits during calculation and round at the end. Let's recalculate:

\( 4.45\times0.046 = 0.2047 \)

\( 0.2047\div5.51 = 0.03715\cdots \)

Now, looking at the significant figures of the given data:

- \( P_1 = 5.51 \, \text{atm} \): three significant figures.
- \( P_2 = 4.45 \, \text{atm} \): three significant figures.
- \( V_2 = 0.046 \, \text{L} \): two significant figures? Wait, no, 0.046: the leading zeros are not significant, so the significant figures are 4 and 6, so two. But wait, maybe 0.046 is considered as two significant figures, but 5.51 and 4.45 are three. However, in the problem statement, when we look at 0.046, maybe it's a typo or maybe I misinterpret. Wait, 0.046 liters: if it's 0.046, that's two significant figures. But 5.51 and 4.45 are three. The rule for multiplication/division is that the result has the same number of significant figures as the number with the least number of significant figures. But here, when we use Boyle's law, \( P_1V_1 = P_2V_2 \), so \( V_1=\frac{P_2V_2}{P_1} \). The values: \( P_2 = 4.45 \) (three), \( V_2 = 0.046 \) (two), \( P_1 = 5.51 \) (three). So the least number of significant figures is two? But our answer came out to approximately 0.037, which is two significant figures? Wait, 0.037 has two significant figures (3 and 7). Wait, no, 0.037: leading zeros are not significant, so 3 and 7, two significant figures. But let's check the calculation again. Wait, maybe 0.046 is three significant figures? No, 0.046: the zeros before the 4 are placeholders, not significant. So 4 and 6 are significant, two. But maybe the problem expects us to consider 0.046 as two significant figures, and 5.51 and 4.45 as three, so the answer should have two? But when we calculated, we got approximately 0.037, which is two significant figures? Wait, 0.037 is two significant figures. Alternatively, maybe 0.046 is considered to have two significant figures, and 5.51 and 4.45 have three, so the answer should be two. But let's check with exact calculation:

\( V_1=\frac{4.45\times0.046}{5.51}=\frac{0.2047}{5.51}\approx0.037 \) (if we take three significant figures, since 5.51 and 4.45 have three, and 0.046 may be considered as two, but maybe the problem considers 0.046 as two, but 5.51 and 4.45 as three, so the limiting factor is two? But 0.037 is two significant figures. So the initial volume is approximately 0.037 liters.

Answer:

0.037 liters