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# A 2.24 L of cylinder of oxygen is found to develop a leakage. When the leakage was plugged the pressure dropped to 570 mm Hg. The number of moles of gas that will escape is : A) 0.025B) 0.050C) 0.075D) 0.09

Last updated date: 20th Jun 2024
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Hint: We can calculate the number of moles of gas that will escape with the help of calculating the drop in pressure and then by putting the values in the ideal gas equation. Now, according to the ideal gas equation,
PV = nRT
Where,
P = pressure of the gas
V = volume of the gas
R = Gas constant
n = number of moles of the gas
T = temperature of the gas

Given :
Initial pressure = 1 atm=760 mm Hg
Final pressure = 570 mm Hg
Hence,
Drop in pressure = 760 - 570 = 190 mm Hg = $\dfrac{190}{760} atm$
Volume = 2.24 L
R = 0.0821 L atm $K^{-1}mol^{-1}$
T=273 K
Now, according to the ideal gas equation,
PV = nRT
or, $n = \dfrac{PV}{RT}$
By putting all the values,
$n= \dfrac{\dfrac{190}{760}\times 2.24}{0.0821\times273}$
= 0.025 mol
Hence, the number of moles of gas that will escape will be 0.025 mol.

So, option (a) is correct.

Note: Most of the time students attempt this question by just putting the given value of pressure in the ideal gas equation, which is wrong. So, we firstly calculate the drop in pressure and put that value of pressure in the ideal gas equation because drop in pressure exactly determines the amount of the gas which has escaped from the cylinder. Also, it should be also kept in mind that the unit of pressure should be the same as that given in the problem.