At 27^{circ} C, a closed vessel contains a mi | vedclass

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(B) Let the weight of each gas be $64 \ g$.The number of moles $(n)$ for each gas is calculated as $n = \frac{	ext{weight}}{	ext{molar mass}}$.For $

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$p_1 > p_2 > p_3$

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(B) Let the weight of each gas be $64 \ g$.The number of moles $(n)$ for each gas is calculated as $n = \frac{	ext{weight}}{	ext{molar mass}}$.For $He$: $n_1 = \frac{64}{4} = 16 \ mol$.For $CH_4$: $n_2 = \frac{64}{16} = 4 \ mol$.For $SO_2$: $n_3 = \frac{64}{64} = 1 \ mol$.Total moles = $16 + 4 + 1 = 21 \ mol$.Partial pressure $p_i = \chi_i 	imes P_{total}$, where $\chi_i$ is the mole fraction.Since $p_i \propto n_i$, the gas with the highest number of moles will have the highest partial pressure.Comparing the moles: $16 (He) > 4 (CH_4) > 1 (SO_2)$.Therefore, $p_1 > p_2 > p_3$.

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