To what temperature should a sample
of gas, be cooled from 25° C to
reduce its volume to half its initial
value at constant pressure.
Answers
Answer:
The answer to this depends on the gas, and at what pressure you're keeping the gas at. An ideal gas is going to follow the ideal gas law (PV=nRT) very closely. If you are unfamiliar with how this law works, you may want to check some of the other answers or some other source. As others have explained using this equation, the temperature at which an ideal gas is going to be half its volume at 25 degrees C (298.15 K) is going to be half the Kelvin temperature. This would be 149.08K which converts to about -124 degrees Celsius. HOWEVER in the real world gasses are not ideal
At temperatures close to the boiling point of a gas, the attraction of the gas's molecules to one another will cause PV to be smaller than the ideal gas law would predict. In other words, you'd have a lower volume for a given pressure and a lower pressure for a given volume. This occurs because there isn't enough kinetic energy for the molecules to impart the repulsive force on themselves and the walls of the container. Putting it another way, as a gas gets closer and closer to being liquid, it's going to behave more an more like a liquid.
An opposite effect occurs when high enough pressures are applied to the gas. At high enough pressure, the size of the molecules themselves becomes a factor we have to consider. The ideal gas law assumes that the actual atoms of a gas have no volume. Which is fine as long as there aren't that many of them because the volume of an atom is incredibly tiny. In a high pressure situation however, all those tiny volumes start to add up, and so for a given temperature, you have a higher PV than you would expect.
These effects are in play even for the noble gasses such as Helium and Neon, though it isn't as pronounced in these gasses. For a more dramatic example of a gas that deviates from the ideal gas law here is a plot of PV/nRT (nRT measured as kJ) for methane at three different temperatures:
(from )
You can see how far these deviations can go from this chart. But -124 isn't on here, so how can we check our answer to this question?
For this we can use van der Waal's equation:
(from )
Here, P, V, n, R, and T all mean the same things as in the Ideal gas law. The new values are a and b, where a relates to the volume of the gas molecules and b relates to the strength of intermolecular interactions. If you set both of these equal to zero you get your PV=nRT equation back.
Assuming we have a mole of methane in a 1 liter container at 22 degrees celsius (V=1,n=1, R=0.08206, T=298.15, a=2.253 L*atm/mol^2, b=0.04278 L/mol), This gives us [P+2.253](1-0.04278)=24.47...
0.9572P-0.09638=24.47.
0.96P=24.56
P=25.59atm
Keeping this pressure constant and halving the volume gives
[25.59+9.012](.5-0.04278)=(0.08206)T
34.60*0.457=(0.08206)T
15.81=(0.08206)T
T=192.69 K or minus 80.45 degrees Celsius.
So for methane, the temperature to halve the volume is more than 42 degrees warmer than the ideal gas law predicts! At higher pressures, this effect would be even more pronounced. And, of course, different gasses are going to have different a values and b values.
BROTHER I BEG YOU BRO PLEASE MARK ME AS A BRAINLEST