If the gas behaves as a perfect gas, it obeys Boyle's Law, which can be stated as 'at constant temperature, pressure is inversely proportional to volume'. In this case the pressure is multiplied by 5/2, so the volume will be 2/5 of the original, i.e. 2 l.
Combined gas law:
P1V1 P2V2
----- = -----
T1 T2
P1= 10.0 ATM
V1= 5.00 L
T1= 500 K
P2= Unknown
V2= 10.0 L
T2= 200 K
10.0 × 5.00 / 500 = P2 × 10.0 / 200
P2= 2.00 ATM
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 = 5.00 L
P2 = Unknown
V2 = 2.00 L
P1V1 = P2V2
1.00(5.00)=2.00P
P= 2.5 ATM
That's Boyle's law. If the temperature remains the same then
(P1)(V1) = (P2)(V2)
where
P1 is the old pressure,
V1 is the old volume,
P2 is the new pressure,
V2 is the new volume.
Substituting:
(200)(2500)=(500)(V2)
500000= (500)(V2)
Divididing both sides by 500
1000 = V2
So the new volume of the gas is 1000 cubic meters.
Note: The gas has been compressed tighter, so it takes up
less space or volume.
Boyle's Law applies to this question. It states that at constant temperature, the volume of a gas is inversely proportional to its pressure. In this case, the volume at 2.0 ATM would be 26 liters.
If the gas behaves as a perfect gas, it obeys Boyle's Law, which can be stated as 'at constant temperature, pressure is inversely proportional to volume'. In this case the pressure is multiplied by 5/2, so the volume will be 2/5 of the original, i.e. 2 l.
P1V1=P2V2
P1=2.0 atm
V1=200 mL
V2=10 mL
P2=(P1V1)/V2
P2=(2.0*200)/(10)
P2=40 atm
Use Boyle's Law, P1V1= P2V2
Rearrange for P2: P2=(P1V1)/V2
P2= (5.0x10.0)/2.0
P2= 25 atm
The pressure of the gas after it is compressed is equal to 25 atmospheres.
1,000 cubic meters..
It is reduced to 2 L.
3800 mm Hg
STP = Standard Temperature and Pressure After the IUPAC rules the standard temperature is 0 0C and the standard pressure is 100 kPa (0,986 atm). The molar volume of an ideal gas at STP is 22,710 980(38) L.
Usually it remains constant. For instance when you boil water and it turns into steam, the temperature of the boiling water remains at 100 Degrees Celsius throughout the process.
Nothing is inincompressible For practical propose, it defined water as incompressible since its' compressibility is very low. To compressed water down to 99/100 of original volume you would need a pressure of 217 Bar approximately. Any normal pressure vessel would burst at such pressure. It is then considered water as incompressible.
You will recall from the Ideal Gas Laws that temperature, pressure, and volume are all connected in terms of the behavior of a gas (especially an ideal gas, but actual gas resembles ideal gas to a certain extent). So, if the gas is in a container of fixed volume, then reducing the temperature will correspondingly reduce the pressure.
240 Assuming ideal gas behavior, doubling the pressure means reducing the volume by a factor of 2.
ice (frozen water) is a tricky material. It melts at 0 degrees C under normal conditions, but if you exceed a pressure of 100 times atmospheric pressure its melting point lowers a few degrees, causing it to melt faster even though temperature is unchanged.
3800 mm Hg
The temperature factor increases to 1.1547, approx.
Corrected Temperature = 100 - [(760mm Hg - Atmospheric Pressure) * (0.037°C/mm)]
It starts from 100 celcius in 1atm and can be risen further. At lower pressures eg on tops of mountains, it can be lower. Water boils when the pressure of water vapour exceeds the external atmospheric pressure. Below that, any bubble of water vapour which might start to form is immediately compressed back to liquid.
no. Compressed nitrogen need not be insulated. Liquid nitrogen is always insulated. Typically compressed nitrogen is stored at high pressure(over 1000 psi) and liquid nitrogen is kept at less than 100 psi.
At a constant 100 C and 1 ATM ambient pressure, I believe it should be 1 as the gaseous and liquid phases would be equally favored. This is due to the fact that the vapor pressure of liquid water at 100 C is 1 ATM.
The temperature of the water is 100 degrees celsius.
Low side pressure is normally 28-38 high side pressure is typically 100 + outside temperature
Use the ideal gas equation to solve this. PV= nRT. You will have to convert your pressure to atmosphere to use the constant R = 0.0821 L*ATM/mol*K. You know your initial pressure, volume, and temperature. Moles can be neglected (n) because they will stay the same. You also know your final pressure and final volume, so you can solve for final temperature.
At standard pressure, it is 100 degrees C.