Question

a balloon filled with carbon dioxide gas (co₂) has a volume of 0.0045 cubic meters and an internal pressure of 1.33 bar when squeezed. releasing the balloon causes the volume of the balloon to increase to 0.0061 cubic meters. what is the final pressure of the co₂ gas in the balloon? assume ideal gas behavior and a constant temperature. write your answer to the correct number of significant figures. round if necessary. bar save answer

Explanation:

Step1: Recall Boyle's Law

Boyle's Law states that for a given mass of an ideal gas at constant temperature, the product of pressure and volume is constant, i.e., \( P_1V_1 = P_2V_2 \).

Step2: Identify known values

We know \( P_1 = 1.33 \) bar, \( V_1 = 0.0045 \) m³, and \( V_2 = 0.0061 \) m³. We need to find \( P_2 \).

Step3: Rearrange the formula to solve for \( P_2 \)

From \( P_1V_1 = P_2V_2 \), we can rearrange to get \( P_2=\frac{P_1V_1}{V_2} \).

Step4: Substitute the known values into the formula

Substitute \( P_1 = 1.33 \), \( V_1 = 0.0045 \), and \( V_2 = 0.0061 \) into the formula:
\( P_2=\frac{1.33\times0.0045}{0.0061} \)
First, calculate the numerator: \( 1.33\times0.0045 = 0.005985 \)
Then, divide by the denominator: \( P_2=\frac{0.005985}{0.0061}\approx0.98114754 \)

Step5: Consider significant figures

The given values \( P_1 = 1.33 \) (3 significant figures), \( V_1 = 0.0045 \) (2 significant figures), \( V_2 = 0.0061 \) (2 significant figures). When multiplying and dividing, the result should have the same number of significant figures as the least precise measurement, which is 2 significant figures? Wait, no: \( V_1 = 0.0045 \) has two significant figures, \( V_2 = 0.0061 \) has two, \( P_1 = 1.33 \) has three. The rule is that the number of significant figures in the result is determined by the least number of significant figures in the values used in the calculation. Here, \( V_1 \) and \( V_2 \) have two significant figures, but \( P_1 \) has three. Wait, actually, \( 0.0045 \) is two significant figures (the leading zeros are not significant, the 4 and 5 are), \( 0.0061 \) is two (6 and 1), and \( 1.33 \) is three. When we do \( P_1V_1 \), \( 1.33\times0.0045 \): 1.33 has three, 0.0045 has two, so the product has two significant figures? Wait, no, the rule for multiplication/division is that the result has the same number of significant figures as the factor with the least number of significant figures. So \( 1.33 \) (3) × \( 0.0045 \) (2) gives a result with 2 significant figures? Wait, no, \( 1.33\times0.0045 = 0.005985 \), if we consider significant figures, 0.0045 has two, so 0.005985 should be rounded to 0.0060 (two significant figures). Then dividing by 0.0061 (two significant figures): \( 0.0060\div0.0061\approx0.98 \). Wait, but maybe I made a mistake here. Let's check the original values again. \( V_1 = 0.0045 \) cubic meters: the trailing zero? No, 0.0045 is 4.5 × 10⁻³, so two significant figures. \( V_2 = 0.0061 \) is 6.1 × 10⁻³, two significant figures. \( P_1 = 1.33 \) is three significant figures. So when we calculate \( P_2=\frac{P_1V_1}{V_2} \), the number of significant figures is determined by the least number, which is two? Wait, no, the formula is \( P_2=\frac{P_1V_1}{V_2} \), so it's (3 sig figs × 2 sig figs) / 2 sig figs. The multiplication of 3 and 2 gives 2 sig figs, then division by 2 sig figs gives 2 sig figs? Wait, no, the rule is that for multiplication and division, the result has the same number of significant figures as the input with the least number of significant figures. So among \( P_1 \) (3), \( V_1 \) (2), \( V_2 \) (2), the least is 2. So the result should have two significant figures? But let's calculate the exact value: \( \frac{1.33\times0.0045}{0.0061}=\frac{0.005985}{0.0061}\approx0.9811 \). Now, 1.33 has three, 0.0045 has two, 0.0061 has two. So the limiting factor is two significant figures? Wait, maybe the problem expects us to use the number of significant figures from the given data. Let's check the original problem: the initial pressure is 1.33 (three sig figs), initial volume 0.0045 (two), final volume 0.0061 (two). So when we do the calculation, the number of sig figs is determined by the least, which is two? But 0.9811 rounded to two significant figures is 0.98? Wait, no, 0.9811 rounded to two significant figures: the first two significant figures are 9 and 8, the next digit is 1, which is less than 5, so it's 0.98? Wait, no, 0.9811: the first significant figure is 9, the second is 8, the third is 1. So two significant figures would be 0.98? Wait, but maybe the problem considers that 0.0045 and 0.0061 have two significant figures, and 1.33 has three, so the result should have two? But let's check with three significant figures: 0.981, which is approximately 0.98 bar? Wait, no, let's do the calculation more accurately. \( 1.33\times0.0045 = 0.005985 \). \( 0.005985\div0.0061 = 0.005985\div0.0061 \). Let's do this division: 0.0061 × 0.98 = 0.005978, which is very close to 0.005985. So 0.98 bar (two significant figures) or 0.981 bar (three)? Wait, the initial pressure is 1.33 (three), initial volume 0.0045 (two), final volume 0.0061 (two). So the rule is that when multiplying/dividing, the result should have the same number of significant figures as the quantity with the least number of significant figures. Here, 0.0045 and 0.0061 have two, so the result should have two. But 0.9811 rounded to two significant figures is 0.98. Wait, but maybe the problem expects three? Let's check the numbers again. 1.33 is three, 0.0045 is two, 0.0061 is two. The formula is \( P_2 = \frac{P_1V_1}{V_2} \). So it's (3 sig figs * 2 sig figs) / 2 sig figs. The 2 sig figs in the numerator (from V1) and 2 in the denominator (V2) would cancel out, leaving 3 sig figs from P1? Wait, no, the multiplication of P1 (3) and V1 (2) gives a number with 2 sig figs, then dividing by V2 (2) gives a number with 2 sig figs. So the answer should be 0.98 bar? Wait, but let's check with the exact calculation: \( \frac{1.33\times0.0045}{0.0061} = \frac{0.005985}{0.0061} \approx 0.98114754 \). Rounded to three significant figures, it's 0.981 bar, to two it's 0.98 bar. But let's see the original values: 1.33 has three, 0.0045 has two, 0.0061 has two. The least number of significant figures is two, so the answer should have two. But maybe the problem considers that 0.0045 and 0.0061 have two, and 1.33 has three, so the result should have two. So the final answer is approximately 0.98 bar? Wait, no, wait: 0.9811 is closer to 0.98 or 0.99? Wait, 0.9811: the third significant figure is 1, so when rounding to two, it's 0.98. But let's check with the initial numbers. Maybe I made a mistake in the significant figures. Let's see: 0.0045 is two sig figs, 0.0061 is two, 1.33 is three. So the formula is \( P_2 = (P_1 \times V_1) / V_2 \). The number of sig figs is determined by the least, which is two. So the answer should be 0.98 bar. Wait, but let's calculate again: \( 1.33 * 0.0045 = 0.005985 \). \( 0.005985 / 0.0061 = 0.98114754 \). So approximately 0.98 bar (two significant figures) or 0.981 bar (three). But the initial pressure is 1.33 (three), so maybe the problem expects three significant figures. Let's check the problem statement: "Write your answer to the correct number of significant figures". The given values: 1.33 (three), 0.0045 (two), 0.0061 (two). The rule is that when multiplying/dividing, the result should have the same number of significant figures as the input with the least number of significant figures. So the least is two, so the answer should have two. But 0.9811 rounded to two significant figures is 0.98. Wait, but 0.9811 is 0.98 when rounded to two significant figures? Wait, no, 0.9811: the first significant figure is 9, the second is 8, the third is 1. So when rounding to two significant figures, we look at the third digit, which is 1, so we keep the second digit as is. So 0.98. But let's check with the exact calculation. Alternatively, maybe the problem considers that 0.0045 and 0.0061 have two significant figures, and 1.33 has three, so the result should have two. So the final answer is 0.98 bar. Wait, but let's see, maybe I made a mistake in the formula. Boyle's Law is \( P_1V_1 = P_2V_2 \) for constant temperature and amount of gas, which is the case here (balloon, so same amount of gas, constant temperature). So the formula is correct. So the calculation is correct. So the final pressure is approximately 0.98 bar.

Answer:

0.98