Let's solve the problem step by step:
### Part (a)
Given:
- Initial pressure, [tex]\( P_1 = 1.8 \times 10^6 \)[/tex] Pascals
- Initial volume, [tex]\( V_1 = 2.1 \times 10^{-8} \)[/tex] m³
- New pressure, [tex]\( P_2 = 4.1 \times 10^4 \)[/tex] Pascals
We need to find the new volume, [tex]\( V_2 \)[/tex].
According to Boyle's Law, at constant temperature, the pressure and volume of a gas are inversely proportional:
[tex]\[ P_1 V_1 = P_2 V_2 \][/tex]
Substituting the given values into the equation:
[tex]\[ (1.8 \times 10^6) \times (2.1 \times 10^{-8}) = (4.1 \times 10^4) \times V_2 \][/tex]
[tex]\[ 3.78 \times 10^{-2} = (4.1 \times 10^4) \times V_2 \][/tex]
Solving for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{3.78 \times 10^{-2}}{4.1 \times 10^4} \][/tex]
[tex]\[ V_2 = \frac{3.78}{4.1} \times 10^{-2 - 4} \][/tex]
[tex]\[ V_2 = 0.92195 \times 10^{-6} \][/tex]
Converting the result to standard form to 3 significant figures:
[tex]\[ V_2 \approx 9.220 \times 10^{-7} \, \text{m}^3 \][/tex]
### Part (b)
Given:
- Initial pressure, [tex]\( P_1 = 1.8 \times 10^6 \)[/tex] Pascals
- Initial volume, [tex]\( V_1 = 2.1 \times 10^{-8} \)[/tex] m³
- New volume, [tex]\( V_2 = 7.1 \times 10^{-9} \)[/tex] m³
We need to find the new pressure, [tex]\( P_2 \)[/tex].
According to Boyle's Law, at constant temperature, the pressure and volume of a gas are inversely proportional:
[tex]\[ P_1 V_1 = P_2 V_2 \][/tex]
Substituting the given values into the equation:
[tex]\[ (1.8 \times 10^6) \times (2.1 \times 10^{-8}) = P_2 \times (7.1 \times 10^{-9}) \][/tex]
[tex]\[ 3.78 \times 10^{-2} = P_2 \times (7.1 \times 10^{-9}) \][/tex]
Solving for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{3.78 \times 10^{-2}}{7.1 \times 10^{-9}} \][/tex]
[tex]\[ P_2 = \frac{3.78}{7.1} \times 10^{-2 + 9} \][/tex]
[tex]\[ P_2 = 0.53239 \times 10^{7} \][/tex]
Converting the result to standard form to 3 significant figures:
[tex]\[ P_2 \approx 5.324 \times 10^{6} \, \text{Pascals} \][/tex]
So, the final answers are:
a) The volume of the gas will be [tex]\( 9.220 \times 10^{-7} \)[/tex] m³.
b) The pressure of the gas will be [tex]\( 5.324 \times 10^6 \)[/tex] Pascals.
