At a constant temperature, the volume of a gas is inversely proportional to its pressure.

Some gas has a pressure of [tex][tex]$1.2 \times 10^6$[/tex][/tex] Pascals and a volume of [tex][tex]$2.3 \times 10^{-8} \, m^3$[/tex][/tex].

a) What will the volume of the gas be when its pressure is [tex][tex]$5.2 \times 10^4$[/tex][/tex] Pascals?

b) Work out the pressure of the gas when its volume is [tex][tex]$4.2 \times 10^{-9} \, m^3$[/tex][/tex].

Give each of your answers in standard form to 3 significant figures.



Answer :

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.