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Why is the ideal gas law usable in most gas problems?

For example, in a problem like this: "Suppose you have $1.00\: \mathrm{mol}$ of a gas at $0\:^\circ \mathrm{C}$, occupying a container which is $500\: \mathrm{mL}$ in size. What is the pressure of this gas in $\mathrm{atm}$?"

The ideal gas law is was applied to solve that problem and the answer was $44.8\: \mathrm{atm}$. But isn't the ideal gas law only applicable to gases which are in high temperatures and low pressure? Because the kinetic energy overcomes the intermolecular bonds and the molar volume is negligible, etc etc.

But the conditions for this "gas" are low temperature and consecutively high pressure which is basically a stamp for real gases so why isn't the real gas law applied to this problem instead?

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    $\begingroup$ There are two key points that make the ideal gas law so universal. One is that it is universal: you don't need to know the actual gas to work with the equation. The second is that there is no single real gas law and the various laws that attempt to improve on the ideal situation require you to know the specific gas involved. Mostly the ideal law is a good approximation, unless you are a chemical engineer working with extreme conditions. $\endgroup$ – matt_black Apr 24 '16 at 12:27
  • $\begingroup$ You have to go to rather arcane substances and/or extreme conditions until the ideal gas law is not correct to within 10%. $\endgroup$ – Karl Mar 30 at 7:09
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You are right in saying that in your question, the gas is an non-ideal (real) gas. Basically, the reason why we use the ideal gas equation despite knowing that it is a real gas is simply because that the equation for real gases is too complicated and contain more variables which are hard to find given limited data.

The reason for this complicated formula is that several assumptions that we make for ideals gases don't hold for real gases. Recall that an ideal gas is considered to be a point mass - a particle so small that the volume of that particle is negligible. A real gas particle does have real volume. For an ideal gas, the collisions between gas particles was said to be "elastic" - no attractive or repulsive forces exist, and thus, no energy is exchanged during collisions. For a real gas, collisions are non-elastic. So, the ideal gas law must be corrected for this extra forces. There are several real gas laws, one is the van der Waals equation:

$$\left[P + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$

Notice how "corrections" are being made to the pressure term and the volume term. Since collisions of real gases are non-elastic, the term $a(n/V)^2$ is correcting for the interactions of these particles. Since real gas particles have real volume, the $nb$ term is correcting for the excluded volume. The values of $a$ and $b$ are constants, and must be determined experimentally for each gas.

As you can see from this equation that first the values of $a$ and $b$ must be known then you must solve this equation which can be complicated. Even then, when you do finally solve it, you will probably get a result which is similar to the result obtained if you used the ideal gas equation. This is why the ideal gas equation is continued to be used as in most cases it provides quite accurate results while bypassing complicated formulas.

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  • $\begingroup$ But I assume that there is another reason for the usage of the ideal gas equation rather than just complicated. Otherwise the answer would be not possible. So let's say that the problem actually gives us the name of the gas. Then I would be able to get a value for the real pressure of the gas using the van der Waal's equation and this value might be close to the value we got using the ideal gas equation. In summary, are you maybe implying that the deviations caused by real gases properties are so small that are not relevant? Which is maybe why the ideal gas law could still be used? $\endgroup$ – Alex Wong Aug 31 '15 at 8:03
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    $\begingroup$ Well, if they did give you the name of the gas you will be able to use the van der Waal's equation. However you are right to say that it will probably give a result that will only deviate slightly form the result obtained from using the ideal gas law. Which is the reason why the ideal has law is used as it still produces relatively accurate results in most cases while bypassing complicated formulas such as the van der Waal's equation. $\endgroup$ – Nanoputian Aug 31 '15 at 8:10
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    $\begingroup$ The other factor here is that assuming the ideal gas law typically allows a solution using a liner or quadratic equation. It isn't really a problem for a computer to solve the van der Waal's equation, but it is awful pain in the neck to do by hand - especially so if you solving for two or three gasses. So using the ideal gas equation let's you apply the chemistry without grinding through a lot of tedious mathematical equations. $\endgroup$ – MaxW Apr 24 '16 at 6:34

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