Question

The weight of $ 350\;mL $ of a diatomic gas at $ 0{}^\circ C $ and $ 2\;atm $ pressure is $ 1\;g $ . The wt. of one atom ?

Answer 1

Hint :Here, we are given the volume of a gas, pressure of the gas, temperature of the gas. Hence, by using universal gas constant with proper units we can find the moles of the gas present in the given weight using the Ideal gas equation. From the moles we can find the number of atoms using the Avogadro number, and as the gas is diatomic, we need to halve the number of atoms to obtain weight of one atom.

Complete Step By Step Answer:
Let us note down the given data and data we need to find;
Volume of the gas (in $ L $ ) $ V=350mL=0.35L $
Pressure of the gas (in $ \;atm $ ) $ p=2atm $
Temperature of gas (in $ K $ ) $ T=0{}^\circ C=273K $
Value of Universal Gas constant (in $ LatmK^{-1}mol^{-1} $ ) $ R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}} $
Moles of the gas (in $ \;mol $ ) $ n=? $
Given mass of the gas (in $ g $ ) $ m=1g $
We have the Ideal Gas equation which can be expressed as
 $ pV=nRT $
As we are interested in finding the moles of the gas, making mole the subject of the equation
 $ \therefore n=\dfrac{pV}{RT} $
Substituting the given values,
 $ \therefore n=\dfrac{2\times 0.35}{0.0821\times 273} $
 $ \therefore n=0.0312mol $
Here, we are given the mass of the gas as $ 1\;g $ . Hence, by equivalency, we can write
 $ \therefore 1g=0.0312mol $
Hence, for $ 1\;mol $ the mass of the gas is
 $ \therefore 1mol=\dfrac{1}{0.0312}=32g $
Here, the gas is diatomic. Hence, the mass of one mole of monatomic gas is
 $ \therefore 1mol=\dfrac{32}{2}g=16g $ …… $ (1) $
Now, we know that Avogadro number shows the number of atoms present in $ 1 $ mole which is
 $ 1mol=6.022\times {{10}^{23}}atoms $ …… $ (2) $
From the above equations,
 $ \therefore 16g=6.022\times {{10}^{23}}atoms $
Hence, the weight per atom will be
 $ \therefore Mass\;of\;one\;atom=\dfrac{16g}{6.022\times {{10}^{23}}atoms} $
 $ \therefore Mass\ of\ one\ atom=2.66\times {{10}^{-23}}gato{{m}^{-1}} $

Note :
The important point we should note here is that the given gas is diatomic. Hence, the Avogadro number will show the number of molecules (not atoms) present in one mole gas. As one molecule contains two atoms of gas, we need to take twice the Avogadro number to show the number of atoms in one mole gas. Also, we should carefully select the value of Universal gas constant by considering the unit in which the pressure and volume are given, as it varies with the units.