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Pressure is halved when ONLY volume is doubled (n and T are constant).Remember the General Gas Law:p.V = n.R.T(in which R=general gas constant)
In a container the volume remain constant but the pressure increase.
The gas takes on the size and shape of the container it's in. So if you make the volume of the container smaller (compress it) the volume of the gas is smaller as well. However, this comes at a higher pressure exerted, so there is no spontaneous mass creation.Well, by definition, compress means "to make smaller; to press or squeeze together; or to make something occupy a smaller space or volume." Therefore, the very word "compress" implies a decrease in volume. So if you wanted to know what happens when you compress a gas, you are squeezing it into a smaller space, or decreasing the volume.If you were to let the gas maintain a constant temperature as you compress it, then pressure would increase. If you were to let the gas maintain a constant pressure, then temperature would decrease.If you were to rephrase your question to "what happens to the volume of gas if put under pressure," then the gas' volume would decrease. For the temperature to remain constant and the pressure to increase, a gas must decrease in volume to occupy a smaller area.
The relationship which is used to answer this is pV=nRT R is a constant, and assuming V and T are also constant, then increasing the number of moles of gas will cause the pressure to increase.
If the volume of the tank was effectivelly constant, and the tank was sealed to prevent gas escaping, the pressure of the gas would increase.
Volume & pressure are inversely proportionate, if temperature stays constant volume would decrease at a factor proporionate to the increase in pressure.
If the temperature of the gas is decreasing, then in order to maintain constant pressure, you would have to compress it in volume.
Pressure is halved when ONLY volume is doubled (n and T are constant).Remember the General Gas Law:p.V = n.R.T(in which R=general gas constant)
It would be -221.7 deg C.
For the pressure to remain the same, the temperature would double if the volume also doubled.
25C is 298K. 52C is 325K. Assuming linearity (an ideal gas), P=3*325/298=3.27 atm.
A sample of Ar gas occupies a volume of 1.2 L at 125°C and a pressure of 1.0 atm. Determine the temperature, in degrees Celsius, at which the volume of the gas would be 1.0 L at the same pressure.
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.
In a container the volume remain constant but the pressure increase.
For gases, there is heat specific heat capacity under the assumption that the volume remains constant, and under the assumption that the pressure remains constant. The reason the values are different is that when heating up a gas, in the case of constant pressure it requires additional energy to expand the gas. For solids and liquids, "constant volume" isn't used, since it would require a huge pressure to maintain the constant volume.
This is the reduction of volume to one-third.
Assuming the tank was not in a vacuum, the VOLUME stays constant. The volume is the total area inside the tank. The pressure would change when 'pumped up'. The volume would not. The pressure inside would also change based on the temperature, relative to the outside pressure.