Wiki User
∙ 13y agoP1V1=P2V2
VI=325ml P1=655mm Hg
V2=125ml P2=?
=655 x 325= P2 x 125
=(655 x 325)/ 125
=212875/125
=1703mm Hg. :)
Wiki User
∙ 13y agoUsing Boyle's law (P1V1 = P2V2), we can determine the new pressure. Plugging in the values: P1 = 655 mm Hg, V1 = 325 mL, V2 = 125 mL, we can solve for P2. The new pressure would be 1706.15 mm Hg.
Some examples of gases that can be compressed are oxygen, carbon dioxide, nitrogen, and helium. When these gases are compressed, their volume decreases while the pressure increases.
Because the helium is compressed.Because that small tank is extreAmly compressed with helium. So it has a LOT of helium in it. It can just fit into the small tank because it's highly compressed.
if its possible then the reason could be the expansion of high pressured helium from tank to balloon is large.
To calculate the volume of compressed air, use the ideal gas law equation: PV = nRT, where P is the pressure of the compressed air, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. This formula allows you to calculate the volume of the compressed air if you know the pressure, temperature, and quantity of air.
A 0.50 mole sample of helium will occupy a volume of 11.2 liters under standard temperature and pressure (STP) conditions, which are 0 degrees Celsius (273.15 K) and 1 atmosphere pressure. At STP, one mole of any gas occupies a volume of 22.4 liters.
Some examples of gases that can be compressed are oxygen, carbon dioxide, nitrogen, and helium. When these gases are compressed, their volume decreases while the pressure increases.
1100
If the pressure on a sample of gas is raised three times and the temperature is kept constant, according to Boyle's Law, the volume of the gas will decrease proportionally to maintain a constant temperature. This means the gas will be compressed and occupy a smaller volume.
Because the helium is compressed.Because that small tank is extreAmly compressed with helium. So it has a LOT of helium in it. It can just fit into the small tank because it's highly compressed.
if its possible then the reason could be the expansion of high pressured helium from tank to balloon is large.
To calculate the volume of compressed air, use the ideal gas law equation: PV = nRT, where P is the pressure of the compressed air, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. This formula allows you to calculate the volume of the compressed air if you know the pressure, temperature, and quantity of air.
A 0.50 mole sample of helium will occupy a volume of 11.2 liters under standard temperature and pressure (STP) conditions, which are 0 degrees Celsius (273.15 K) and 1 atmosphere pressure. At STP, one mole of any gas occupies a volume of 22.4 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 = 1.30 ATM V1 = 31.4 L P2 = Unknown V2 = 15.0 L P1V1 = P2V2 1.30(31.4)=15.0P P= 2.72 ATM Looking at the question simply, you can get an estimate on the pressure because you can see that pressure and volume vary indirectly (if volume goes up, pressure goes down). If the volume is cut in half (roughly), then the pressure should increase by half.
When a gas is compressed, the gas particles are pushed closer together, reducing the space between them. This results in a decrease in volume due to the increased pressure and decrease in container size.
When a gas is compressed, its volume decreases while the number of gas molecules remains the same. This leads to the gas particles being more concentrated in a smaller space, resulting in an increase in pressure according to the ideal gas law, which states that pressure is inversely proportional to volume.
To solve for the original pressure of the helium gas, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Using this law, we can set up the equation (P1)(V1) = (P2)(V2), where P1 is the original pressure, V1 is the original volume, P2 is the final pressure, and V2 is the final volume. Plugging in the values gives us (P1)(200 mL) = (300 mm Hg)(0.240 mL). Solving for P1 gives us P1 = (300 mm Hg)(0.240 mL) / 200 mL = 0.36 mm Hg. Therefore, the original pressure of the helium gas was 0.36 mm Hg.
Are you stating or asking ? If that's a statement, then it's an incorrect one. At constant temperature, the product of (pressure) x (volume) is constant. So, if the volume changed by a factor of 3, the pressure must also change by a factor of 3 ... the pressure must triple.