The air in the syringe is at a pressure of 2.0×105Pa. The piston is slowly moved into the
syringe, keeping the temperature constant, until the volume of the air is reduced from
80 cm3 to 25 cm3. Calculate the final pressure of the air.
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.