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

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(A) From the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.Rearranging the equation,we get $\frac{P

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

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(A) From the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.Rearranging the equation,we get $\frac{PV}{T} = \frac{mR}{M}$.Since the masses $m$ are equal and $R$ is a constant,we have $\frac{PV}{T} \propto \frac{1}{M}$.From the graph,the slope of the lines is $\frac{PV}{T}$. The slope is greatest for line $C$ and smallest for line $A$,so $(\frac{PV}{T})_C > (\frac{PV}{T})_B > (\frac{PV}{T})_A$.This implies $M_C The molar masses are $M_{H_2} = 2 	ext{ g/mol}$,$M_{He} = 4 	ext{ g/mol}$,and $M_{O_2} = 32 	ext{ g/mol}$.Therefore,$M_{H_2} Comparing the two relations,we get $C$ corresponds to $H_2$,$B$ to $He$,and $A$ to $O_2$.

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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