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

Vedclass · Accepted Answer

$M_{1} > M_{2}$

Vedclass · Answer

(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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