The PV versus T graph for equal masses of H_2 | vedclass

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(D) From the ideal gas equation,$PV = \mu RT$,where $\mu$ is the number of moles.Rearranging this,we get $PV = (\mu R)T$.Comparing this with the equat

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$A$ corresponds to $O_2, B$ to $He$ and $C$ to $H_2$

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(D) From the ideal gas equation,$PV = \mu RT$,where $\mu$ is the number of moles.Rearranging this,we get $PV = (\mu R)T$.Comparing this with the equation of a straight line $y = mx$,where $y = PV$ and $x = T$,the slope of the graph is $m = \mu R$.Since $R$ is a universal gas constant,the slope is directly proportional to the number of moles $\mu$.For a given mass $m$,the number of moles is $\mu = \frac{m}{M}$,where $M$ is the molar mass.Thus,$\mu \propto \frac{1}{M}$.The molar masses are $M(H_2) = 2 	ext{ g/mol}$,$M(He) = 4 	ext{ g/mol}$,and $M(O_2) = 32 	ext{ g/mol}$.Since $M(H_2)  \mu(He) > \mu(O_2)$.Consequently,the slopes follow the order $	ext{Slope}(H_2) > 	ext{Slope}(He) > 	ext{Slope}(O_2)$.Looking at the graph,the slope of line $C$ is the greatest,followed by $B$,and then $A$.Therefore,$C$ corresponds to $H_2$,$B$ corresponds to $He$,and $A$ corresponds to $O_2$.This matches option $D$.

The $PV$ versus $T$ graph for equal masses of $H_2, He$,and $O_2$ is shown in the figure. Choose the correct alternative:

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