4 g of an ideal gas A (molar mass = M_A) pre | vedclass

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(D) Using the ideal gas equation $PV = nRT = \frac{wRT}{M}$, we have $n = \frac{PV}{RT}$.For gas $A$: $n_A = \frac{5V}{RT} = \frac{4}{M_A} \implies M_

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$4 M_A = M_B$

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(D) Using the ideal gas equation $PV = nRT = \frac{wRT}{M}$, we have $n = \frac{PV}{RT}$.For gas $A$: $n_A = \frac{5V}{RT} = \frac{4}{M_A} \implies M_A = \frac{4RT}{5V}$.When gas $B$ is added, the total pressure becomes $10 \ atm$. The partial pressure of gas $B$ is $P_B = P_{total} - P_A = 10 \ atm - 5 \ atm = 5 \ atm$.For gas $B$: $n_B = \frac{5V}{RT} = \frac{16}{M_B} \implies M_B = \frac{16RT}{5V}$.Taking the ratio: $\frac{M_A}{M_B} = \frac{4RT/5V}{16RT/5V} = \frac{4}{16} = \frac{1}{4}$.Therefore, $4 M_A = M_B$.

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