Answer:[tex]P_2=2.5\times 10^5\ Pa[/tex]
Explanation:
Given
Initial volume [tex]V_1=10\times 10^{-6}\ m^3[/tex]
Initial Pressure [tex]P_1=10^5\ Pa[/tex]
Final trapped air [tex]V_2=4\times 10^{-6}\ m^3[/tex]
If there is no change in temperature then We can write
[tex]P_1V_1=P_2V_2\quad \text{(From ideal gas equation)}[/tex]
[tex]P_2=\dfrac{P_1V_1}{V_2}[/tex]
[tex]P_2=10^5\times \frac{10\times 10^{-6}}{4\times 10^{-6}}[/tex]
[tex]P_2=2.5\times 10^5\ Pa[/tex]
