Question

A cylinder containing 5 L of $O_2$ at $25{}^{\circ }\text{C}$ was leaking. When the leakage was detected and stopped, there was a change in the pressure of the gas from 3 atm to 2.235 atm. What volume of oxygen got leaked at 1 atm and $25{}^{\circ }\text{C}$?

• A. 4.8 L
• B. 3.8 L
• C. 2.8 L
• D. 5.8 L

Answer 1

The correct option is B 3.8 L

At $P_1=3.0$ atm
$n_1=\frac{P_{1}V}{RT}=\frac{3V}{RT}$
At $P_2=2.235$ atm
$n_2=2.235$ atm
$n_2=\frac{P_{2}V}{RT}=\frac{2.235\times V}{RT}$
$\therefore$ No. of moles leaked $n=n_1-n_2$
$n=\frac{3V}{RT}-\frac{2.235V}{RT}$
$=0.765\frac{V}{RT}$
$=\frac{0.765\times 5}{0.0821\times 298}$
$=0.156$ mol
$\therefore$ Volume of $O_2$ leaked $=\frac{nRT}{P}=\frac{0.156\times 0.0821\times 298}{1}=3.8\text{L}$