Question

An open vessel is heated from $\text{ 2}{{\text{7}}^{\text{0}}}\text{C }$ to$\text{ 22}{{\text{7}}^{\text{0}}}\text{C }$. The percentage of air expelled out is:
(A) 20
(B) 30
(C) 40
(D) 50

Answer 1

Hint: The ideal gas equation is as shown below,
$\text{ PV = nRT }$
Where the P is the pressure of the gas, R is the gas constant, V is the volume of the gas, T is the absolute temperature and n is the number of moles of gas.
The gas in the open vessel is heated, then on applying the ideal gas equation the pressure and the volume on the vessel will remain constant.

Complete step by step solution:
The ideal gas equation establishes the relation between the pressure of the gas (P), the volume of gas (V), number of moles of gas (n), the absolute temperature of the gas (T), and the gas constant.
The ideal gas is given as follows,
$\text{ PV }\propto \text{ nT}$
That is, $\text{ PV = nRT }$
Where,
P is the pressure, V is the volume of gas, n is the number of moles, R is the gas constant and T is the absolute temperature.
We have provided the following data:
Temperature $\text{ (}{{\text{T}}_{\text{1}}}\text{) }$ is equal to,
$\text{ }{{\text{T}}_{\text{1}}}\text{ = 2}{{\text{7}}^{\text{0}}}\text{C = 27 + 273 K = 300 K}$
The vessel is heated to the temperature $\text{ (}{{\text{T}}_{2}}\text{) }$is equal to,
$\text{ }{{\text{T}}_{2}}\text{ = 22}{{\text{7}}^{\text{0}}}\text{C = 227 + 273 K = 500 K}$
We have to find the percentage of air expelled from the vessel. The percentage of the gas expelled from the vessel can be calculated as follows,
During the heating of the vessel, the pressure and volume of the gas remain constant. Therefore ,the ideal gas equation at the temperature $\text{ (}{{\text{T}}_{\text{1}}}\text{) }$and $\text{ (}{{\text{T}}_{2}}\text{) }$is written as follows,
$\text{ }{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{ = }{{\text{n}}_{\text{1}}}\text{R}{{\text{T}}_{\text{1}}}\text{ }$ And $\text{ }{{\text{P}}_{2}}{{\text{V}}_{2}}\text{ = }{{\text{n}}_{2}}\text{R}{{\text{T}}_{2}}\text{ }$
Or $\text{ }{{\text{V}}_{\text{1}}}\text{ = }\dfrac{{{\text{n}}_{\text{1}}}\text{R}{{\text{T}}_{\text{1}}}}{{{\text{P}}_{\text{1}}}}\text{ }$and $\text{ }{{\text{V}}_{2}}\text{ = }\dfrac{{{\text{n}}_{2}}\text{R}{{\text{T}}_{2}}\text{ }}{{{\text{P}}_{2}}}\text{ }$
The gas is heated in the same vessel the number of moles, volume, and pressure remain the same. Since the pressure and volume is the same the number of moles of the gas at temperature $\text{ (}{{\text{T}}_{\text{1}}}\text{) }$and$\text{ (}{{\text{T}}_{2}}\text{) }$, the volume of the vessel will remain the same therefore we have,
$\begin{align}
& \text{ }\dfrac{{{\text{n}}_{\text{1}}}\text{}{{\text{T}}_{\text{1}}}}{{\text{}}}\text{ = }\dfrac{{{\text{n}}_{2}}\text{}{{\text{T}}_{2}}\text{ }}{{\text{}}} \\
& \Rightarrow \text{ }{{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }{{\text{n}}_{2}}{{\text{T}}_{2}} \\
\end{align}$
Let's substitute the values in the equation. We have,
$\begin{align}
& \text{ }{{\text{n}}_{\text{1}}}\text{ (300}\text{.15 ) = }{{\text{n}}_{2}}\text{ (500}\text{.15 )} \\
& \text{ }{{\text{n}}_{2}}\text{ = }\dfrac{\text{300}}{\text{500}}\text{ }{{\text{n}}_{\text{1}}} \\
& \therefore \text{ }{{\text{n}}_{2}}\text{ = }\dfrac{3}{5}{{\text{n}}_{\text{1}}} \\
\end{align}$
$\text{ }{{\text{n}}_{2}}\text{ }$Are the moles of gas which are present at the temperature $\text{ (}{{\text{T}}_{2}}\text{) }$.therefore, the moles of gas which are expelled out would be equal to the,$\text{ }{{\text{n}}_{1}}\text{= 1}-\dfrac{3}{5}{{\text{n}}_{\text{1}}}\text{ = }\dfrac{2}{5}{{\text{n}}_{\text{1}}}$
We can also write it as, \[\text{ moles of gas expelled in }{\scriptstyle{}^{0}/{}_{0}}\text{ = }\dfrac{2}{5}{{\text{n}}_{\text{1}}}\text{ }\times \text{ }\dfrac{100}{{{\text{n}}_{\text{1}}}}\text{ = 40}{\scriptstyle{}^{0}/{}_{0}}\text{ }\]
Therefore, the number of moles of gas expelled from the vessel is equal to the \[\text{ 40}{\scriptstyle{}^{0}/{}_{0}}\] of the moles of the total gas.

Hence, (C) is the correct option.

Note: The ideal gas is a combination of the kinetic gas equation:
1) Boyle’s law: for a definite mass of gas and at the constant temperature, the $\text{ PV = constant }$
2) Charles’s law: for a definite quantity of gas (constant N) the volume of a gas varies with the temperature (T), $\text{ V }\propto \text{ T }$
3) Avogadro’s law: The equal volumes of the gases under the same condition of temperature and pressure have an equal number of the molecules. $\text{ V }\propto \text{ n }$
Therefore, $\text{ V }\propto \text{ }\dfrac{\text{nT}}{\text{P}}\text{ }$