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Question: A \(2.25l\) of a cylinder of oxygen at \(\text{0}{}^\circ \text{C}\) and \(1\) atm is found to devel...

A 2.25l2.25l of a cylinder of oxygen at 0∘C\text{0}{}^\circ \text{C} and 11 atm is found to develop a leakage. When the leakage was plugged, the pressure dropped to 570570 mm of Hg. The number of moles of gas that escaped will be?
(A) 0.025
(B) 0.050
(C) 0.075
(D) 0.09

Explanation

Solution

For the given problem we have to use the gas equation that is PV = nRT where P is the pressure, V is the volume of gas, n is the number of moles of gas, R is known as gas constant and T is the temperature.

Complete step by step solution:
-In the given question, we have to find the number of moles of gas it is given that the volume of a cylinder, temperature and pressure dropped of the gas is 2.25l2.25l, 273K273K and 570570 mm of Hg respectively.
-As we know that the gas equation is given by PV = nRT\text{PV = nRT} where P is the pressure, V is the volume of gas, n is the number of moles of gas, R is known as gas constant and T is the temperature.
-By using the gas equation and putting all the given values we can calculate the number of moles of the gas i.e.
PV = nRT\text{PV = nRT} or it can be written as
n = PVRT\text{n = }\dfrac{\text{PV}}{\text{RT}}
-So, as we know that the value of V is given 2.25l2.25l and the T temperature is given in Celsius we will convert it into kelvin i.e.
0∘C = 0 + 273 = 273K0{}^\circ \text{C = 0 + 273 = 273K} and we knew that the value of R, gas constant is fixed that is 0.0821 L atm K−1 mol−1\text{0}\text{.0821 L atm }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}.
-Also, we know that the value of pressure at NTP is 760 mm of Hg and the final pressure is 570 mm of Hg, the total pressure will be:
Pinitial - Pfinal = 760 - 570 = 190/760 mm of Hg{{\text{P}}_{\text{initial}}}\text{ - }{{\text{P}}_{\text{final}}}\text{ = 760 - 570 = 190/760 mm of Hg}
-So, the value of the number of moles will be:
n = 190 × 2.25760 × 0.0821 × 273 = 0.025 mol.\text{n = }\dfrac{190\text{ }\times \text{ 2}\text{.25}}{760\,\times \text{ 0}\text{.0821 }\times \text{ 273}}\text{ = 0}\text{.025 mol}\text{.}

Therefore, option (A) is the correct answer.

Note: Ideal gas equation gives us the relationship between pressure, volume, temperature, number of moles and gas constant. But the real gas equation is different from it that is (P + an2V2) (V - nb) = nRT\text{(P + }\dfrac{\text{a}{{\text{n}}^{2}}}{{{\text{V}}^{2}}})\text{ (V - nb) = nRT} where a and b are the constant.