V1 = 4L, P1 = 2.07 atm, P2= ?, V2 = 2.5L
We will use following equation to solve this problem: P1V1/T1 = P2V2/T2
Assume Temperature is constant (as nothing is mentioned about temperature in our problem) so our new equation will be P1V1 = P2V2
Plugging in the values (2.07 atm)(4L) = P2(2.5L)
so P2 = (2.07atm) (4L)/(2.5L) (Rearranging the terms)
P2 = 3.312 atm
Pressure and volume are inversely related at constant temperature.
If the same temperature is maintained, then the new volume is 2.056 L. (rounded)
If you double the pressure, the volume will be reduced by the same factor - 4 liters in this case.
Pressure and temperature. As pressure increases, volume decreases; as temperature increases, volume increases with it. At standard temperature and pressure (1 atm, 273 degrees Kelvin), one mole of a gas (6.022 x 1023 particles) has the volume of 22.4 liters.
It would be -221.7 deg C.
For this you would use Boyle's Law, P1V1 = P2V2. The first pressure and volume variables are before the change, while the second set are after the change. In this case, the volume is being changed and the pressure has to be solved for. P1 = 1.00 ATM V1 = 2.0 L P2 = Unknown V2 = 4.00 L P1V1 = P2V2 1.00(2.0)=4.00P P= .5 ATM
This pressure, of nearly one atm., is basically the atmospheric pressure which exists at sea level.
1.1
A sample of gas occupies 1.55L at STP. What will the volume be if the pressure is increased to 50 atm while the temperature remains constant?
5.3 liters
Decreasing the pressure of a gas will increase its volume -- C
2.79 ATM
The ideal gas law states that PV = nRT, where P = pressure and V = volume. Assuming that nRT are all constant, then V = nRT/P. If P decreases by 1/2, then V will double, leaving a final volume of 4 L.
If 125 ml of He gas at 100 degrees C and .0981 atm is cooled to 25 degrees C and the pressure increased to 1.15 atm the new volume is 104 ml.
1.75 atm
What is the volume of a 24.7 mol gas sample that has a pressure of .999 ATM at 305 K?
false
condensationCondensation- Apex
The water is in the gas phase.