Question

A 22.4 L cylinder of oxygen at 1 atm and 273 K is found to develop a leakage. When the leakage was plugged the pressure dropped to 570mm of Hg. The number of moles of gas that escaped will be:
A. 0.244 mol
B. 0.050 mol
C. 0.075 mol
D. 0.09 mol

Answer 1

Hint: Recall the ideal gas equation and all the quantities like pressure, volume, number of moles, and temperature that it relates. Think about which of the quantities are constant in this problem and which are varying. We can find the answer using the ideal gas equation.

Complete step by step answer:
- The ideal gas equation is defined as the product of the pressure and volume being equal to the product of the number of moles and the temperature. The constant of proportionality used is the universal gas constant. So, the equation in mathematical form is as follows:
\[PV=nRT\]
Where all the variables take their usual meanings.

- We know that for the given question the volume of oxygen gas ($V=22.4L$) and temperature of gas ($T=273K$) remains constant. So, pressure of oxygen gas is directly proportional to its number of moles i.e., $P\text{ }\alpha \text{ }n$ where $P$ is the pressure of gas and $n$ is the number of moles of oxygen.
- We will have to convert the value of pressure given in terms of mm of Hg into atm for easy calculation. For this, remember that 1 atm = 760 mm of Hg

- First, we will have to find the change in pressure that occurs as per the given values. The value that we get of the number of moles when we plug in the value for the change in pressure will be the change in the number of moles inside the container. So, the modified equation will be:
\[(\Delta P)V=(\Delta n)RT\]

- Now, to find $\Delta P$ we will have to subtract the final pressure from the initial pressure whose values are given as 570 mm of Hg and 1 atm respectively. We have to convert the value given in mm of Hg into atm since the universal gas constant that we are going to use will have the units $Latm/molK$ or $L\text{ }atm\text{ }mo{{l}^{-1}}{{K}^{-1}}$.

- So, the final pressure in atm by cross multiplication will be:
\[\begin{align}
 & \dfrac{1\text{ atm}}{760\text{ mm of Hg}}=\dfrac{\text{final pressure in atm}}{\text{570 mm of Hg}} \\
 & \text{final pressure in atm}=\dfrac{570}{760} \\
 & \text{final pressure in atm}=0.75atm \\
\end{align}\]
Now, subtracting the final pressure from the initial pressure to obtain $\Delta P$.
\[\begin{align}
 & \Delta P=1atm-0.75atm \\
 & \Delta P=0.25atm \\
\end{align}\]

- Now, we will put this value along with all the other values in the modified equation of the ideal gas, and solve for $\Delta n$.
\[\begin{align}
 & 0.25\times 22.4=\Delta n\times 0.082\times 273 \\
 & \Delta n=\dfrac{5.6}{22.386} \\
 & \Delta n=0.250 \\
 & \Delta n\cong 0.244 \\
\end{align}\]
So, the correct answer is “Option A”.

Note: Note that equal volume of all gases under the same conditions of temperature and pressure contains equal number of molecules. Also, remember that is the volume and the number of moles of a substance are constant, the pressure of a given gas is directly proportional to its temperature in degrees Kelvin.