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Why $ 22.4litres $ called the molar volume of gas?

seo-qna
Last updated date: 23rd Jul 2024
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Answer
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Hint :Molar volume of a gas is defined as the volume of one mole of the gas. Thus, the molar volume is also the volume occupied by $ 6.02{\text{ }} \times {\text{ }}{10^{23}}\; $ particles of gas . $ STP $ refers to the standard temperature of $ {0^ \circ }C $ and pressure of $ 1{\text{ }}atm\; $ . Room conditions refer to the temperature of $ 25^\circ C $ and the pressure of $ 1{\text{ }}atm\; $ .

Complete Step By Step Answer:
Always the Standard temperature and pressure $ \left( {STP} \right) $ is defined as $ 0^\circ C{\text{ }}\left( {273.15{\text{ }}K} \right) $ and $ 1{\text{ }}atm $ pressure. The molar volume of a gas is the volume of one mole of a gas at $ STP $ . At $ STP $ one mole
 ( $ 6.02{\text{ }} \times {\text{ }}{10^{23}}\; $ representative particles) of any gas occupies a volume of $ 22.4{\text{ }}L $ . Standard Molar Volume is the volume occupied by one mole of any gas at $ STP $ .
The Molar Volume of an ideal gas at $ STP $ , which we define to be $ $ $ {0^ \circ }C $ and $ 1{\text{ }}atm\; $ arbitrarily (because we're old-fashioned and stuck in 1982) is $ 22.411{\text{ }}L/mol. $
To calculate this we can use the Ideal gas law of $ PV = nRT $ At $ STP $ (Standard Temperature and Pressure), we CHOSE:
 $ P = 1{\text{ }}atm $
 $ V = ? $
 $ n = 1{\text{ }}mol $
 $ R = 0.082057{\text{ }}L \cdot atm/mol \cdot K $
 $ T{\text{ }} = {\text{ }}273.15{\text{ }}K \\
    \\ $
 $ V = nRTP $
  $ = (1mol)(0.082057atm \cdot Lmol \cdot K)273.15K1atm $
= $ 22.4{\text{ }}L $
This is the volume of one mole of ideal gas at STP.

Note :
We have not specified the identity of the gas; we have specified only that the pressure is $ 1{\text{ }}atm\; $ and the temperature is $ 273{\text{ }}K $ . This makes for a very useful approximation: any gas at $ STP $ has a volume of $ 22.4{\text{ }}L $ per mole of gas; that is, the molar volume at $ STP $ is $ 22.411{\text{ }}L/mol. $ .This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at $ STP $ . If the conditions are not at $ STP $ , a molar volume of $ 22.411{\text{ }}L/mol. $ is not applicable. However, if the conditions are not at $ STP $ , the combined gas law can be used to calculate the volume of the gas at $ STP $ ; then the $ 22.411{\text{ }}L/mol. $ molar volume can be used.