A vessel contains 3.2 g of dioxygen gas at STP (273.15 K and 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in L is: (Given - molar volume at STP is 22.4 L )

Question

A vessel contains 3.2 g of dioxygen gas at STP (273.15 K and 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in L is: (Given - molar volume at STP is 22.4 L ) (A) 6.72 (B) 2.24 (C) 22.4 (D) 67.2

Tardigrade Admin · Accepted Answer

Moles of Dioxygen gas =(3.2 g /32 g mol -1)=0.1 mol volume of Dioxygen gas at STP =0.1 × 22.4 =2.24 L At constant temperature P 1 V 1= P 2 V 2 1 × 2.24=(1/3) × V 2 V 2=6.72 L

Q. A vessel contains $3.2 g$ of dioxygen gas at STP ( $273.15 K$ and $1 a t m$ pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in $L$ is : (Given - molar volume at STP is $22.4 L$ )

$6.72$

$2.24$

$22.4$

$67.2$

Moles of Dioxygen gas $= \frac{3.2 g}{32 g m o l ^{- 1}} = 0.1 m o l$
volume of Dioxygen gas at $STP = 0.1 \times 22.4$
$= 2.24 L$
At constant temperature
$P_{1} V_{1} = P_{2} V_{2}$
$1 \times 2.24 = \frac{1}{3} \times V_{2}$
$V_{2} = 6.72 L$

Q. A vessel contains 3.2g of dioxygen gas at STP (273.15K and 1atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in L is : (Given - molar volume at STP is 22.4L )

6.72

2.24

22.4

67.2