Answer: 1 gram of [tex]H_2[/tex].
Explanation: Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.
[tex]V\propto n[/tex] (At constant temperature and pressure)
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas
[tex]V_2[/tex] = final volume of gas
[tex]n_1[/tex] = initial number of moles
[tex]n_2[/tex] = final number of moles
a) 1 gram of [tex]O_2[/tex]
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{1g}{32g/mol}=0.03125moles[/tex]
c) 1 gram of Ar
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{1g}{40g/mol}=0.025moles[/tex]
d) 1 gram of [tex]H_2[/tex]
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{1g}{2g/mol}=0.5moles[/tex]
Thus the one having highest number of moles will have highest volume which is for 1 gram of [tex]H_2[/tex].
