The air in the syringe is at a pressure of 2.0×105Pa. The piston is slowly moved into the

P V = n R T

n R and T do not change so

P1 V1 = P2 V2

2 *10^5 * 80 = P2 * 25
so
P2 = 2*10^5 * (80/25)

The air in the syringe is at a pressure of 2.0 X10^5 Pa. The piston is slowly moved into the syringe,

keeping the temperature constant, until the volume of the air is reduced from 80 cm^3 to 25 cm^3.

Calculate the final pressure of the air

E3

To calculate the final pressure of the air in the syringe, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature

In this case, the temperature is constant, so we can assume that T1 = T2. Therefore, the equation becomes:

P1V1 = P2V2

Let's plug in the given values:
P1 = 2.0×10^5 Pa
V1 = 80 cm^3 = 80 × 10^(-6) m^3
V2 = 25 cm^3 = 25 × 10^(-6) m^3

Now we can solve for P2:

P2 = (P1 × V1) / V2

P2 = (2.0 × 10^5 Pa × 80 × 10^(-6) m^3) / (25 × 10^(-6) m^3)

P2 = (2.0 × 80) × 10^5 / 25

P2 = 160 × 10^5 / 25

P2 = 6.4 × 10^6 Pa

Therefore, the final pressure of the air in the syringe is 6.4 × 10^6 Pa.