At a pressure 43 kPa, the gas in a cylinder has a volume of 11 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 5 liters, what is the new pressure?
kPa
Im guessing it 58 or 77x10^3
PV=kT
so, since T is constant, so is PV. You want P such that
P*5 = 43*11
P = 94.6
To find the new pressure, we can use the principle of Boyle's Law, which states that the pressure and volume of a gas are inversely proportional.
Boyle's Law can be expressed as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Given:
Initial pressure (P1) = 43 kPa
Initial volume (V1) = 11 liters
Final volume (V2) = 5 liters (the volume decreased)
Now we can substitute the values into the equation:
P1V1 = P2V2
Solving for P2:
P2 = (P1V1) / V2
P2 = (43 kPa * 11 liters) / 5 liters
P2 = 473 kPa
Therefore, the new pressure of the gas, when the volume is decreased to 5 liters, is 473 kPa.