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* 10^5\ Pa`

Step-by-step explanation:

Given

Initial volume
`V_1=10* 10^(-6)\ m^3`

Initial Pressure
`P_1=10^5\ Pa`

Final trapped air
`V_2=4* 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=(P_1V_1)/(V_2)`

`P_2=10^5* (10* 10^(-6))/(4* 10^(-6))`

`P_2=2.5* 10^5\ Pa`