# How can the ideal gas law be made more precise?

##### 1 Answer

**ASSUMPTION OF THE IDEAL GAS LAW**

#PV = nRT => \mathbf(PbarV = RT)#

The **ideal gas law** really assumes that all gases at STP have a **molar volume**

*But in real life, some gases are easier to compress than a typical ideal gas, and some are harder to compress. i.e. the molar volume will not always be that of the ideal gas.*

**"MODIFYING" THE IDEAL GAS LAW TO MAKE IT MORE REALISTIC**

The **pressure** and **temperature** can be measured in real life with good certainty, and the **universal gas constant** is well-known to at least 6 decimal places, so

The **experimental mass density** tends to be well-known for many substances.

Thus, to account for non-ideality, use the *experimental mass density* of the gas, and *convert* to the molar volume via the molar mass. That corrects for the *only* variable in the ideal gas law that assumes ideality.

**EXAMPLE OF USING THE MASS DENSITY**

Let's say we looked at calculating the molar volume for **helium**. Assuming ideality, we simply claim about

Moving along, let's use its mass density (acquired from the equation on pg. 9) to see what we get.

#rho = "0.17614"# #(pm 0.00005)# #"g/L"# at#0^@ "C"# and#"1 bar"#

Since **relative atomic mass**

#color(blue)(barV_"real") = (M_r)/rho#

#= ("4.002602" cancel"g""/mol")/("0.17614" cancel"g""/L")#

#=# #color(blue)("22.724 L")#

So, this means helium, having a *slightly* higher molar volume than that of the ideal gas, is *slightly* **harder to compress** than an ideal gas if treated realistically.

**Ultimately, this can be interpreted as near-ideality at STP.**

**ALTERNATIVE: COMPRESSIBILITY FACTOR**

Another way you could have done it is to introduce the **compressibility factor**

#color(green)(PbarV = ZRT)# ,where

#Z = (PbarV)/(RT)# indicates whether a gas iseasierto compress than the ideal gas (#Z < 1# ),harderto compress than the ideal gas (#Z > 1# ), orideal(#Z = 1# ).

But that can be harder to look up quickly, as it is dependent on the reduced temperature and pressure as per the Theorem of Corresponding States, so personally I find using the density more convenient.

I found it using this equation (pg. 9) to be **slightly harder to compress** than an ideal gas, as stated above.

As before, we'd get:

#color(blue)(barV_"real") = (ZRT)/P = (1.0005*"0.083145 L"cdotcancel"bar""/mol"cdotcancel("K")*"273.15" cancel"K")/(cancel"1 bar")#

#= "22.722 L" ~~ color(blue)("22.724 L")# (within#0.009%# error... close enough.)