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.
To find the new volume of the gas, you can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature is constant. Using the formula P1V1 = P2V2, you can calculate the new volume by rearranging the equation to V2 = (P1)(V1) / P2, where P1 = initial pressure, V1 = initial volume, and P2 = final pressure. Plug in the values to find the new volume.
If the same temperature is maintained, then the new volume is 2.056 L. (rounded)
The volume of the gas will decrease proportionally to the increase in pressure, following Boyle's Law. Using the formula P1V1 = P2V2, where P1 = 12 ATM, V1 = 23 L, and P2 = 14 ATM, we can solve for V2 to find the new volume of the gas. Solving for V2 gives V2 = (P1)(V1) / P2 = (12)(23) / 14 = 19.71 liters.
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
water changes from a gas to a solid to a liquid
To experience a pressure of 2 ATM, you would need to dive to a depth of 20 meters (2 ATM = 1 ATM (surface) + 1 ATM (pressure at 10 meters depth)). At a depth of 100 meters, the pressure would be approximately 11 ATM (1 ATM at surface + 1 ATM for every 10 meters).
This pressure, of nearly one atm., is basically the atmospheric pressure which exists at sea level.
1.1
Using the ideal gas law equation, we can calculate the new volume of the gas. At STP, the pressure is 1 atm, which means 50 atm is 50 times greater. So the new volume would be 1.55L / 50 = 0.031L, when the pressure is increased to 50 atm.
5.3 liters
A. An increase in pressure from 2 ATM to 3 ATM will result in a decrease in volume of gas. B. An increase in pressure from 3 ATM to 4 ATM will result in a decrease in volume of gas. C. A decrease in pressure from 4 ATM to 1 ATM will result in an increase in volume of gas. D. An increase in pressure from 1 ATM to 3 ATM will result in a decrease in volume of gas.
2.79 ATM
As pressure increases from 0.8 ATM to 1.2 ATM, the boiling point of water also increases. Therefore, at 100Β°C and 0.8 ATM, water would boil, but at 1.2 ATM, the water would need to be heated to a higher temperature to reach the new boiling point under the increased pressure.
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 the pressure of a gas increases from 2 atm to 3 atm at constant temperature, the new volume will decrease by a factor equal to the initial pressure divided by the final pressure (V2 = V1 * P1/P2). So, the new volume would be 2/3 times the initial volume.
false
condensationCondensation- Apex
If the oxygen is used at standard pressure (1 ATM), the volume of oxygen available will be 5.0 liters. This is because the volume of a gas is directly proportional to its pressure when the temperature remains constant, according to Boyle's Law (P1V1 = P2V2).
The volume is 3,81 L.