For the same mass of two different ideal gase | vedclass

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(A) For an ideal gas,the ideal gas equation is $pV = \frac{w}{M}RT$,where $w$ is the mass,$M$ is the molecular weight,$R$ is the gas constant,and $T$

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$M_{1} > M_{2}$

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(A) For an ideal gas,the ideal gas equation is $pV = \frac{w}{M}RT$,where $w$ is the mass,$M$ is the molecular weight,$R$ is the gas constant,and $T$ is the temperature.Since $w$,$R$,and $T$ are constant,let $wRT = K$. Then $pV = \frac{K}{M}$,or $V = \frac{K}{Mp}$.Taking the logarithm on both sides: $\log V = \log \left( \frac{K}{M} \right) - \log p$.This is in the form of $y = mx + c$,where $y = \log V$,$x = \log p$,$m = -1$,and the intercept $c = \log \left( \frac{K}{M} \right)$.From the graph,the intercept for $M_{2}$ is higher than the intercept for $M_{1}$.Therefore,$\log \left( \frac{K}{M_{2}} \right) > \log \left( \frac{K}{M_{1}} \right)$.This implies $\frac{K}{M_{2}} > \frac{K}{M_{1}}$,which means $M_{1} > M_{2}$.

For the same mass of two different ideal gases of molecular weights $M_{1}$ and $M_{2}$,plots of $\log V$ vs $\log p$ at a given constant temperature are shown. Identify the correct option.

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