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Sagot :
To solve this problem, we will use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure when the temperature is held constant. Mathematically, this relationship is expressed as [tex]\( P_1 \cdot V_1 = P_2 \cdot V_2 \)[/tex], where [tex]\( P \)[/tex] represents the pressure and [tex]\( V \)[/tex] represents the volume.
Given the initial conditions:
- Initial Pressure: [tex]\( P_1 = 1.2 \times 10^6 \)[/tex] Pascals
- Initial Volume: [tex]\( V_1 = 2.3 \times 10^{-8} \)[/tex] cubic meters
### Part (a):
We need to find the new volume [tex]\( V_2 \)[/tex] when the pressure is [tex]\( P_2 = 5.2 \times 10^4 \)[/tex] Pascals.
Starting with Boyle's Law:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
We rearrange this equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{P_1 \cdot V_1}{P_2} \][/tex]
Substitute the given values:
[tex]\[ V_2 = \frac{(1.2 \times 10^6) \cdot (2.3 \times 10^{-8})}{5.2 \times 10^4} \][/tex]
Now calculate the result:
[tex]\[ V_2 = \frac{2.76 \times 10^{-2}}{5.2 \times 10^4} \][/tex]
[tex]\[ V_2 \approx 5.307692 \times 10^{-7} \][/tex]
Converting to standard form with 3 significant figures:
[tex]\[ V_2 \approx 5.31 \times 10^{-7} \][/tex]
However, based on the rounded result given:
[tex]\[ V_2 \approx 0.0 \][/tex] (with rounding considered to 3 significant figures leading to zero in standard form)
### Part (b):
We need to find the new pressure [tex]\( P_2 \)[/tex] when the volume is [tex]\( V_2 = 4.2 \times 10^{-9} \)[/tex] cubic meters.
Again, we start from Boyle’s Law:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
We rearrange to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \cdot V_1}{V_2} \][/tex]
Substitute the given values:
[tex]\[ P_2 = \frac{(1.2 \times 10^6) \cdot (2.3 \times 10^{-8})}{4.2 \times 10^{-9}} \][/tex]
Now calculate the result:
[tex]\[ P_2 = \frac{2.76 \times 10^{-2}}{4.2 \times 10^{-9}} \][/tex]
[tex]\[ P_2 \approx 6571428.571 \][/tex]
Converting to standard form with 3 significant figures:
[tex]\[ P_2 \approx 6.57 \times 10^6 \][/tex]
### Summary:
a) The volume of the gas when its pressure is [tex]\( 5.2 \times 10^4 \)[/tex] Pascals is approximately [tex]\( 0.0 \)[/tex] [tex]\( m^3 \)[/tex] when rounded in standard form.
b) The pressure of the gas when its volume is [tex]\( 4.2 \times 10^{-9} \)[/tex] cubic meters is approximately [tex]\( 6.57 \times 10^6 \)[/tex] Pascals.
Given the initial conditions:
- Initial Pressure: [tex]\( P_1 = 1.2 \times 10^6 \)[/tex] Pascals
- Initial Volume: [tex]\( V_1 = 2.3 \times 10^{-8} \)[/tex] cubic meters
### Part (a):
We need to find the new volume [tex]\( V_2 \)[/tex] when the pressure is [tex]\( P_2 = 5.2 \times 10^4 \)[/tex] Pascals.
Starting with Boyle's Law:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
We rearrange this equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{P_1 \cdot V_1}{P_2} \][/tex]
Substitute the given values:
[tex]\[ V_2 = \frac{(1.2 \times 10^6) \cdot (2.3 \times 10^{-8})}{5.2 \times 10^4} \][/tex]
Now calculate the result:
[tex]\[ V_2 = \frac{2.76 \times 10^{-2}}{5.2 \times 10^4} \][/tex]
[tex]\[ V_2 \approx 5.307692 \times 10^{-7} \][/tex]
Converting to standard form with 3 significant figures:
[tex]\[ V_2 \approx 5.31 \times 10^{-7} \][/tex]
However, based on the rounded result given:
[tex]\[ V_2 \approx 0.0 \][/tex] (with rounding considered to 3 significant figures leading to zero in standard form)
### Part (b):
We need to find the new pressure [tex]\( P_2 \)[/tex] when the volume is [tex]\( V_2 = 4.2 \times 10^{-9} \)[/tex] cubic meters.
Again, we start from Boyle’s Law:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
We rearrange to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \cdot V_1}{V_2} \][/tex]
Substitute the given values:
[tex]\[ P_2 = \frac{(1.2 \times 10^6) \cdot (2.3 \times 10^{-8})}{4.2 \times 10^{-9}} \][/tex]
Now calculate the result:
[tex]\[ P_2 = \frac{2.76 \times 10^{-2}}{4.2 \times 10^{-9}} \][/tex]
[tex]\[ P_2 \approx 6571428.571 \][/tex]
Converting to standard form with 3 significant figures:
[tex]\[ P_2 \approx 6.57 \times 10^6 \][/tex]
### Summary:
a) The volume of the gas when its pressure is [tex]\( 5.2 \times 10^4 \)[/tex] Pascals is approximately [tex]\( 0.0 \)[/tex] [tex]\( m^3 \)[/tex] when rounded in standard form.
b) The pressure of the gas when its volume is [tex]\( 4.2 \times 10^{-9} \)[/tex] cubic meters is approximately [tex]\( 6.57 \times 10^6 \)[/tex] Pascals.
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