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A cylinder of unknown volume contain $H e (g)$ at $3.5$ atm and $315 K$ . The helium gas is then transferred to a $7 L$ gas cylinder containing $N e (g)$ , at $2.5$ atm and $315 K$ . If the final total pressure at $315 K$ is $5.5$ atm, then what is the volume of cylinder (in $L$ ) that initially contained the $H e (g)$ .

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Detailed Solution

Let volume of $H e$ cylinder be $V L$

$n_{H e} = \frac{(3.5 V) L a t m}{315 R} & n_{N e} = \frac{(2.5 \times 7) L a t m}{315 R} (5.5 \times 7) L . a t m = \frac{(3.5 V + 2.5 \times 7) L a t m}{315 R} \times 315 R 3.5 V = (5.5 - 2.5) \times 7 \Rightarrow V = \frac{3 \times 7}{3.5} = 6$

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