A vessel contains $1.6 g$ of dioxygen at STP $(273.15 K, 1 a t m$ pressure ). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate the volume of the new vessel.

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Solution

$p_{1} = 1 a t m, T_{1} = 273 K, V_{1} = ?$
$32 g$ of oxygen occupies $22.4 L$ of volume of STP
Hence, $1.6 g$ of $O_{2}$ will occupy,
$1.6 g O_{2} \times \frac{22.4 L}{32 g O_{2}} = 1.12 L$
$V_{1} = 1.12 L$
$p_{2} = \frac{p_{1}}{2} = \frac{1}{2} = 0.5 a t m$
$V_{2} = ?$
According to Boyle's law:
$p_{1} V_{1} = p_{2} V_{2}$
$V_{2} = \frac{p_{1} \times V_{1}}{p_{2}} = \frac{1 a t m \times 1.12 L}{0.5 a t m}$
$= 2.24 L$

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A vessel contains 1.6 g of dioxygen at STP (273.15 K,1 atm pressure ). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate the volume of the new vessel.

p1=1 atm,T1=273 K,V1=?32 g of oxygen occupies 22.4 L of volume of STPHence, 1.6 g of O2 will occupy,1.6 g O2×22.4 L32 gO2=1.12 LV1=1.12 Lp2=p12=12=0.5 atmV2=?According to Boyle's law:p1V1=p2V2V2=p1×V1p2=1 atm×1.12 L0.5 atm=2.24 L

A vessel contains $1.6 g$ of dioxygen at STP $(273.15 K, 1 a t m$ pressure ). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate the volume of the new vessel.