If an ideal gas is allowed to expand into a vacuum, this means that the external pressure is 0. This doesn't affect the internal pressure though, correct?
For example:
I have a problem in which one mole of an ideal gas at 300. K and at a volume of 10.0 L expands isothermally into a vacuum, which doubles its volume. If I plugged these values into an equation, I would have:
0.00 atm x 10.0 L= 1.00 x .08206 x 300. L
Which doesn't work. So the pressure used in this equation is not 0, correct?
well the volume doubles, so it's now 20L. Since the problem provided you with everything but the correct pressure, you would be able to calculate the pressure by using PV=nRT. I calculated 1.24 atm or 936 torr.
you might want to double check me though. It's been a while since I've done ideal gases. :)
You are correct that when an ideal gas expands into a vacuum, the external pressure is indeed zero. However, this does not mean that the internal pressure of the gas is also zero.
In the example you provided, one mole of an ideal gas at 300 K and at a volume of 10.0 L expands isothermally into a vacuum and doubles its volume to 20.0 L. To calculate the pressure, you can use the ideal gas law equation, PV = nRT.
First, let's determine the values we know:
- P = pressure (unknown)
- V = volume (initially 10.0 L, then doubles to 20.0 L)
- n = moles of gas (1 mole)
- R = ideal gas constant (0.08206 L.atm/mol.K)
- T = temperature (300 K)
Now, let's plug in these values into the ideal gas law equation:
P * 10.0 L = 1.0 mole * 0.08206 L.atm/mol.K * 300 K
Simplifying the equation:
P = (1.0 mole * 0.08206 L.atm/mol.K * 300 K) / 10.0 L
Calculating the value:
P = 2.4618 atm
Therefore, the pressure of the gas after it expands into a vacuum and doubles its volume is approximately 2.46 atm or 936 torr.
Remember, the vacuum outside the system has no effect on the internal pressure of the gas, which is determined by its temperature, volume, and the number of moles present.