At 273,K,Pd vs P is plotted for various gases | vedclass

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(D) From the ideal gas equation,$PV = nRT = (\frac{m}{M})RT$. Rearranging this,$P(\frac{m}{V}) = \frac{PRT}{M}$,which gives $Pd = \frac{PM}{RT}$. For

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$2-N_2, 3-He, 1-CO_2, 4-H_2$

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(D) From the ideal gas equation,$PV = nRT = (\frac{m}{M})RT$. Rearranging this,$P(\frac{m}{V}) = \frac{PRT}{M}$,which gives $Pd = \frac{PM}{RT}$. For a given temperature $T$,the slope of the $Pd$ vs $P$ graph is $\frac{M}{RT}$,which is directly proportional to the molar mass $M$ of the gas. The molar masses are: $CO_2 = 44\,g/mol$,$N_2 = 28\,g/mol$,$He = 4\,g/mol$,$H_2 = 2\,g/mol$. Since the slope increases as molar mass increases,the order of slopes is $CO_2 > N_2 > He > H_2$. Looking at the graph,the slope follows the order $1 > 2 > 3 > 4$. Therefore,$1-CO_2, 2-N_2, 3-He, 4-H_2$. Matching this with the given options,option $D$ is $2-N_2, 3-He, 1-CO_2, 4-H_2$.

At $273\,K$,$Pd$ vs $P$ is plotted for various gases $1, 2, 3, 4$. Assuming ideal behaviour for four gases $N_2, He, CO_2$ & $H_2$. The correct combination is [$P$ denotes pressure in atmosphere & $d$ denotes density in $g/L$]

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