Question

An evacuated bulb of known volume is filled with $H_2$ gas at room temperature $\left({30}^{o}C\right)$. The pressure of the gas in the bulb is 750 mm Hg. A portion of the gas is transferred to a different flask and found to occupy a volume of 50.0 mL at 1 atm pressure and at the same temperature. The pressure of the $H_2$ gas remaining in the original bulb drops to 600 mm Hg. What is the volume of the bulb assuming $H_2$ gas is an ideal gas ?

Answer 1

For ideal gas, equation of state is given by $PV=nRT$

Let $V$ be the volume of bulb and $n_1$ moles are there initially then we have the initial state equation as,

$756\times V=n_{1}RT⟶\left(I\right)$

let $n_2$ moles are transferred to flask, then we have

$760\times 40=n_2\times RT⟶\left(II\right)$

Now we have remaining $(n_1-n_2)$ moles in the bulb at pressure $625$ m of $Hg$.

Hence we have,

$625\times V=(n_1-n_2)RT=n_{1}RT-n_2\times R\times T⟶\left(III\right)$

by substituting for $n_{1}RT$ from $\left(I\right)$ and for $n_{2}RT$ from $\left(II\right)$ and in equation $\left(III\right)$, we can solve the equation $\left(III\right)$ we get, $V=232ml$