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Question: The weight of \( 350\;mL \) of a diatomic gas at \( 0{}^\circ C \) and \( 2\;atm \) pressure is \( 1...

The weight of 350  mL350\;mL of a diatomic gas at 0C0{}^\circ C and 2  atm2\;atm pressure is 1  g1\;g . The wt. of one atom ?

Explanation

Solution

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 LL ) V=350mL=0.35LV=350mL=0.35L
Pressure of the gas (in   atm\;atm ) p=2atmp=2atm
Temperature of gas (in KK ) T=0C=273KT=0{}^\circ C=273K
Value of Universal Gas constant (in LatmK1mol1LatmK^{-1}mol^{-1} ) R=0.0821LatmK1mol1R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}}
Moles of the gas (in   mol\;mol ) n=?n=?
Given mass of the gas (in gg ) m=1gm=1g
We have the Ideal Gas equation which can be expressed as
pV=nRTpV=nRT
As we are interested in finding the moles of the gas, making mole the subject of the equation
n=pVRT\therefore n=\dfrac{pV}{RT}
Substituting the given values,
n=2×0.350.0821×273\therefore n=\dfrac{2\times 0.35}{0.0821\times 273}
n=0.0312mol\therefore n=0.0312mol
Here, we are given the mass of the gas as 1  g1\;g . Hence, by equivalency, we can write
1g=0.0312mol\therefore 1g=0.0312mol
Hence, for 1  mol1\;mol the mass of the gas is
1mol=10.0312=32g\therefore 1mol=\dfrac{1}{0.0312}=32g
Here, the gas is diatomic. Hence, the mass of one mole of monatomic gas is
1mol=322g=16g\therefore 1mol=\dfrac{32}{2}g=16g …… (1)(1)
Now, we know that Avogadro number shows the number of atoms present in 11 mole which is
1mol=6.022×1023atoms1mol=6.022\times {{10}^{23}}atoms …… (2)(2)
From the above equations,
16g=6.022×1023atoms\therefore 16g=6.022\times {{10}^{23}}atoms
Hence, the weight per atom will be
Mass  of  one  atom=16g6.022×1023atoms\therefore Mass\;of\;one\;atom=\dfrac{16g}{6.022\times {{10}^{23}}atoms}
Mass of one atom=2.66×1023gatom1\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.