Answer
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Hint: The molar volume of a gas is the volume of one mole of a gas at STP. At STP, temperature is taken as 273 K and pressure is taken as 1atm. We can use the equation of ideal gas to calculate the volume of 1 mole of gas.
Complete answer:
At STP, one mole (\[6.02\times {{10}^{23}}\] representative particles) of any gas occupies a volume of 22.4 L . A mole of any gas occupies 22.4L at standard temperature and pressure (273K and 1atm).
\[PV=nRT\]
where \[P=1\] atom, \[R=0.0821\] L atom \[mo{{l}^{-1}}{{K}^{-1}},\text{ }T=273K\text{ }n=1\], mol
Substituting these values in the above equation we get,
\[1\times V=1\times 0.08201\times 273\]
\[V=1\times 0.08201\times 273\]
\[V=22.4L\].
Hence, the correct answer is true which is option A.
Note: The above statement was a law given by Avogadro. It is an experimental gas law relating the volume of a gas to the amount of substance of gas present in that particular volume. This law is a specific case of the ideal gas law. Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of particles.". Thus, all gases will occupy a volume of 22.4 L at 273 K and 1 atm pressure.
Complete answer:
At STP, one mole (\[6.02\times {{10}^{23}}\] representative particles) of any gas occupies a volume of 22.4 L . A mole of any gas occupies 22.4L at standard temperature and pressure (273K and 1atm).
\[PV=nRT\]
where \[P=1\] atom, \[R=0.0821\] L atom \[mo{{l}^{-1}}{{K}^{-1}},\text{ }T=273K\text{ }n=1\], mol
Substituting these values in the above equation we get,
\[1\times V=1\times 0.08201\times 273\]
\[V=1\times 0.08201\times 273\]
\[V=22.4L\].
Hence, the correct answer is true which is option A.
Note: The above statement was a law given by Avogadro. It is an experimental gas law relating the volume of a gas to the amount of substance of gas present in that particular volume. This law is a specific case of the ideal gas law. Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of particles.". Thus, all gases will occupy a volume of 22.4 L at 273 K and 1 atm pressure.
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