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(B) Let $n$ be the total number of moles of gas in both bulbs initially. Since the bulbs have identical volumes $V$ and are at the same temperature $T

Vedclass · Accepted Answer

$2 / 3$

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(B) Let $n$ be the total number of moles of gas in both bulbs initially. Since the bulbs have identical volumes $V$ and are at the same temperature $T$,the number of moles in each bulb is $n_1 = n_2 = n / 2$.When bulb $2$ is heated to $2 T$ and bulb $1$ is at $T$,the pressure $p$ in both bulbs must be equal for the system to be in equilibrium.Using the ideal gas law $p V = n R T$,the number of moles in bulb $1$ $(n_1')$ and bulb $2$ $(n_2')$ are:$n_1' = \frac{p V}{R T}$$n_2' = \frac{p V}{R (2 T)} = \frac{p V}{2 R T}$Since the total number of moles is conserved,$n_1' + n_2' = n = n_1 + n_2 = n / 2 + n / 2 = n$.$\frac{p V}{R T} + \frac{p V}{2 R T} = n \Rightarrow \frac{3 p V}{2 R T} = n \Rightarrow \frac{p V}{R T} = \frac{2 n}{3}$.Thus,the final number of moles in bulb $2$ is $n_2' = \frac{1}{2} 	imes \frac{p V}{R T} = \frac{1}{2} 	imes \frac{2 n}{3} = \frac{n}{3}$.The ratio of the final mass (or final moles) in bulb $2$ to the initial mass (or initial moles) in bulb $2$ is:Ratio $= \frac{n_2'}{n_2} = \frac{n / 3}{n / 2} = \frac{2}{3}$.

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