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Question: Molar‌ ‌volume‌ ‌is‌ ‌the‌ ‌volume‌ ‌occupied‌ ‌by‌ ‌1‌ ‌mol‌ ‌of‌ ‌any‌ ‌(idea)‌ ‌gas‌ ‌at‌ ‌standa...

Molar‌ ‌volume‌ ‌is‌ ‌the‌ ‌volume‌ ‌occupied‌ ‌by‌ ‌1‌ ‌mol‌ ‌of‌ ‌any‌ ‌(idea)‌ ‌gas‌ ‌at‌ ‌standard‌ ‌temperature‌ ‌and‌ ‌pressure‌ ‌(S‌ ‌T‌ ‌P:‌ ‌1‌ ‌atmospheric‌ ‌pressure,‌ ‌00C{{0}^{0}}\,C).‌ ‌Show‌ ‌that‌ ‌it‌ ‌is‌ ‌22.4‌ ‌litres.‌ ‌

Explanation

Solution

In order to answer the following question, we need to will acquainted with the features of modern periodic table and the following formula the maximum number of electrons present in shell = 2n22{{n}^{2}}
Where n = shell number
To prove that the molar volume at standard pressure and temperature is 224.4 litres can be proved using the ideal gas law PV=nRtPV=nRt,where R is the universal gas constant

Complete step-by-step answer: It is given to us that Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard pressure and temperature (S T P: 1 atmosphere pressure, =OoC={{O}^{o}}C )
As it is given in the question, molar volume is basically the volume occupied by 1 mole of any (ideal) gas at standard pressure and temperature.
So, from it we can conclude that the standard pressure is 1 atmosphere (ATM) while standard temperature is 0oC{{0}^{o}}C
Now, in order to show that the molar volume at S T P is 22.4 litres we will use the ideal gas equation.
Now, we know that the ideal gas equation relating pressure (P), volume (V) and absolute temperature (T) is given as follows-
PV=nRTPV=nRT
Where 'R' is the universal gas constant with value R=8.314Jmol1K1R=8.314Jmo{{l}^{-1}}{{K}^{-1}}
n = number of moles
here, n = 1
T = standard temperature
=OoC={{O}^{o}}C
= 273 K
P = standard pressure = 1 atm = 1.013×105Nm2\text{P = standard pressure = 1 atm = 1}\text{.013}\times \text{1}{{\text{0}}^{5}}N{{m}^{-2}}
In order to find the volume we need to rearrange the ideal gas equation as depicted below.
Thus, we can write
V=nRT/PV=nRT/P
Now substituting the values of n,R,T,Pn,\,R,\,T,\,P in the above equation and solving it we will get volume as,
V=1×8.314×2731.013×105\Rightarrow \,V\,=\dfrac{1\times 8.314\times 273}{1.013\times {{10}^{5}}}
V=0.0224m3\therefore V=0.0224{{m}^{3}}
Now, we know that
Im3=1000 Liters{{\operatorname{Im}}^{3}}=1000\text{ Liters}
So on converting the meter cube unit to liters we get,
V=0.224×1000 litres\therefore V=0.224\times 1000\text{ litres}
V=22.4V=22.4 liters
Hence, We can see that the molar volume of a gas at S T P is 22.4 litres. Therefore, it is proved that one mole of a gas occupies 22.4L22.4L of the volume at STP.

Note: Students should note that for real gases the molar volume at S T P does not equal to 22.4 litres. Which leads us to questioning the accuracy of PV=nRTPV=nRT and this can only be judged by comparing the actual volume of 1 mole of gas to the molar this is given by the compressibility factor.