Question

A sealed syringe contains 10 × 10−6 m3 of air at 1 × 105 Pa. The plunger is pushed until the volume of trapped air is 4 × 10−6 m3. If there is no change in temperature, what is the new pressure of the gas?

Answer 1

Answer:$P_2=2.5\times 10^5\ Pa$

Explanation:

Given

Initial volume $V_1=10\times 10^{-6}\ m^3$

Initial Pressure $P_1=10^5\ Pa$

Final trapped air $V_2=4\times 10^{-6}\ m^3$

If there is no change in temperature then We can write

$P_1V_1=P_2V_2\quad \text{(From ideal gas equation)}$

$P_2=\dfrac{P_1V_1}{V_2}$

$P_2=10^5\times \frac{10\times 10^{-6}}{4\times 10^{-6}}$

$P_2=2.5\times 10^5\ Pa$