What is the volume of $1 m o l e$ of gas particles at STP?

22.4 L

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22.2 L

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42.2 L

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10 L

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Solution

The correct option is A $22.4 L$
$P V = n R T$ $-$ Ideal gas law
given, $P = 1$ atm
$T = 273.15$ K
$n = 1$
$V = \frac{n R T}{P}$
$V = \frac{1 \times 0.082 \times 273.15}{1}$
$V = 22.4$ L

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What is the volume of 1 mole of gas particles at STP?

The correct option is A 22.4LPV=nRT − Ideal gas lawgiven, P=1 atm T=273.15 K n=1V=nRTPV=1×0.082×273.151V=22.4 L

What is the volume of $1 m o l e$ of gas particles at STP?

The correct option is A $22.4 L$
$P V = n R T$ $-$ Ideal gas law
given, $P = 1$ atm
$T = 273.15$ K
$n = 1$
$V = \frac{n R T}{P}$
$V = \frac{1 \times 0.082 \times 273.15}{1}$
$V = 22.4$ L