Answer:
the final pressure in pascal = [tex]6.30*10^5 \ \ Pa[/tex]
Explanation:
Given that:
initial pressure = [tex]6.9*10^5 \ Pa[/tex]
Initial Temperature = 17.5 °C = (17.5 + 273) K = 290.5 K
Volume = 2.00 L = 2.00 × 10⁻³ m³
Initial volume of the air occupied by gas V' = 95 cm³ = 95× 10⁻⁶ m
Using the ideal gas temperature;
PV = nKT
where
K = Boltzmann constant = 1.38 × 10⁻²³
From above expression;
[tex]n = \frac{PV}{KT}[/tex]
[tex]n = \frac{6.9*10^5*2.00*10^{-3}}{1.38*10^{-23}*290.5}[/tex]
[tex]n = 3.44*10^{23}[/tex]
The number of moles that were removed from the tire is calculated as
[tex]\Delta \ n = \frac{P_{atm}*V'}{KT}[/tex]
where
[tex]P_{atm}[/tex] = atmospheric pressure = [tex]1.013*10^5[/tex]
[tex]\Delta \ n = \frac{1.013*10^5*95*10^{-6}}{1.38*10^{-23}*290.5}[/tex]
[tex]\Delta \ n = 2.4*10^{21}[/tex]
The remaining number of moles after the release of gas is
[tex]n_2 = n- \Delta n[/tex]
[tex]n_2 = 3.44*10^{23} -2.4*10^{21}[/tex]
[tex]n_2 = 3.416*10^{23}[/tex]
Using the expression [tex]P_2 = \frac{n_2 KT}{V}[/tex] to determine the final pressure:
[tex]P_2 = \frac{3.416*10^{23}*1.38*10^{-23}*290.5 }{2.00*10^{-3}}[/tex]
[tex]P_2 = 6.30*10^5 \ \ Pa[/tex]
Hence, the final pressure in pascal = [tex]6.30*10^5 \ \ Pa[/tex]
