Question

A $2.25l$ of a cylinder of oxygen at $\text{0}{}^\circ \text{C}$ and $1$ atm is found to develop a leakage. When the leakage was plugged, the pressure dropped to $570$ 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

Answer 1

Hint: 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 $2.25l$, $273K$ and $570$ mm of Hg respectively.
-As we know that the gas equation is given by $\text{PV = nRT}$ 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.
$\text{PV = nRT}$ or it can be written as
$\text{n = }\dfrac{\text{PV}}{\text{RT}}$
-So, as we know that the value of V is given $2.25l$ and the T temperature is given in Celsius we will convert it into kelvin i.e.
$0{}^\circ \text{C = 0 + 273 = 273K}$ and we knew that the value of R, gas constant is fixed that is $\text{0}\text{.0821 L atm }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}$.
-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:
${{\text{P}}_{\text{initial}}}\text{ - }{{\text{P}}_{\text{final}}}\text{ = 760 - 570 = 190/760 mm of Hg}$
-So, the value of the number of moles will be:
$\text{n = }\dfrac{190\text{ }\times \text{ 2}\text{.25}}{760\,\times \text{ 0}\text{.0821 }\times \text{ 273}}\text{ = 0}\text{.025 mol}\text{.}$

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 $\text{(P + }\dfrac{\text{a}{{\text{n}}^{2}}}{{{\text{V}}^{2}}})\text{ (V - nb) = nRT}$ where a and b are the constant.