A container contains 32 ,g of O_2 at a temper | vedclass

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(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.For the first container: $m(O_2) = 32 \,g$,$M(O_2) =

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$4P$

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(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.For the first container: $m(O_2) = 32 \,g$,$M(O_2) = 32 \,g/mol$. Thus,$n_1 = \frac{32}{32} = 1 \,mol$.The pressure is $P = \frac{n_1 RT}{V} = \frac{1 \cdot RT}{V} = \frac{RT}{V}$.For the second container: $m(H_2) = 4 \,g$,$M(H_2) = 2 \,g/mol$. Thus,$n_2 = \frac{4}{2} = 2 \,mol$.The temperature is $2T$. The pressure $P'$ is given by $P' = \frac{n_2 R(2T)}{V} = \frac{2 \cdot R(2T)}{V} = \frac{4RT}{V}$.Substituting $\frac{RT}{V} = P$,we get $P' = 4P$.

$A$ container contains $32 \,g$ of $O_2$ at a temperature $T$. The pressure of the gas is $P$. An identical container containing $4 \,g$ of $H_2$ at a temperature $2T$ has a pressure of ............

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