Two vessels A and B contain oxygen. The volum | vedclass

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(B) From the ideal gas equation,$PV = nRT = \frac{N}{N_A} RT$,where $N$ is the number of molecules and $N_A$ is Avogadro's number.Thus,$N = \frac{PVN_

Vedclass · Accepted Answer

$1:12$

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(B) From the ideal gas equation,$PV = nRT = \frac{N}{N_A} RT$,where $N$ is the number of molecules and $N_A$ is Avogadro's number.Thus,$N = \frac{PVN_A}{RT}$.For vessel $A$: $N_A = \frac{p_A V_A N_A}{R T_A}$.For vessel $B$: $N_B = \frac{p_B V_B N_A}{R T_B}$.Given: $V_B = 2V_A$,$p_B = 3p_A$,and $T_B = \frac{T_A}{2}$.Taking the ratio $\frac{N_A}{N_B}$:$\frac{N_A}{N_B} = \frac{p_A V_A}{R T_A} \cdot \frac{R T_B}{p_B V_B} = \frac{p_A V_A}{T_A} \cdot \frac{T_A / 2}{3p_A \cdot 2V_A} = \frac{1}{T_A} \cdot \frac{T_A}{2 \cdot 3 \cdot 2} = \frac{1}{12}$.Therefore,the ratio $N_A : N_B = 1:12$.

Two vessels $A$ and $B$ contain oxygen. The volume of $B$ is twice that of $A$,the pressure of $B$ is thrice that of $A$ and the temperature of $B$ is half of that of $A$. Then,find the ratio of number of molecules of oxygen in vessels $A$ and $B$.

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