Question

An open container of volume ${\text{V}}$contains air at temperature $27^\circ {\text{C}}$ or ${\text{300K}}$ . The container is heated to such a temperature so that amount of gas coming out is $\dfrac{{\text{2}}}{{\text{3}}}$ of
Amount of gas initially present in the container.
Amount of gas finally remaining in the container.
Find the temperature to which the container should be heated.

Answer 1

To solve this question, checkout at the parameters given to us within the question. Since Pressure-${\text{(P)}}$.Volume-${\text{(V)}}$ and real gas constant-${\text{(R)}}$ remains constant, relate the number of moles and temperature by using the ideal gas equation.
Formula used:
${\text{PV = nRT}}$
where P is the pressure, V is the volume, n is the moles, R is the gas constant and T is the temperature.

Complete step by step answer:
According to the question, there’s an open vessel with a temperature equal to $27^\circ {\text{C}}$ or ${\text{300K}}$ .
Also, two-third of air escapes. Therefore, we will say that there’s a change in moles of the gas.
So, let the initial moles in air be ${{\text{n}}_{\text{1}}}$and therefore the number of moles after two-third of gas escaped be${{\text{n}}_{\text{2}}}$.
Let ${{\text{n}}_{\text{1}}}$= $1$mole
So,
${{\text{n}}_{\text{2}}}$ = ${{\text{n}}_1} - \dfrac{2}{3} = 1 - \dfrac{2}{3} = \dfrac{1}{3}$ moles.
According to the question, we will say that the volume remains constant.
Ideal gas equation relates ${\text{PV = nRT}}$
Since, Pressure-${\text{(P)}}$.Volume-${\text{(V)}}$and real gas constant-${\text{(R)}}$remain constant.
We will relate the number of moles and temperature as ${\text{nT}}$= constant
${{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}{\text{ = }}{{\text{n}}_{\text{2}}}{{\text{T}}_{\text{2}}}$=constant
Now, putting the value of moles and temperature we get
$1(300) = \left( {\dfrac{1}{3}} \right){{\text{T}}_{\text{2}}}$
${{\text{T}}_{\text{2}}} = \dfrac{{300 \times 3}}{1}{\text{K}}$
Solving this, we get:
${{\text{T}}_{\text{2}}} = 900{\text{K}}$
Let there be ${\text{x}}$ moles of the gas remaining with in the container , $\dfrac{2}{3}$ of ${\text{x}}$comes out
$\therefore \left( {\dfrac{2}{3}{\text{x}} + {\text{x}}} \right)$ = ${\text{n}}$ = $\dfrac{{5{\text{x}}}}{3}$ = ${\text{n}}$
$\therefore {\text{x}} = \dfrac{{3{\text{n}}}}{5}$
Using ${{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}{\text{ = }}{{\text{n}}_{\text{2}}}{{\text{T}}_{\text{2}}}$
${\text{n}} \times (300K) = (\dfrac{{{\text{3n}}}}{5}){{\text{T}}_2}$
Now, cancel ${\text{n}}$on each side and we will get ${{\text{T}}_2}$
$\therefore {{\text{T}}_2} = 500{\text{K}}$

Hence final Temperature =$500{\text{K}}$

Note:
Ideal gas equation is the equation of state of an ideal gas (hypothetical). It is an approximation of the behavior of gases under ideal conditions. lt is a combination of empirical laws like Boyle's law, Charles law, Gay-Lussac's law, and Avogadro's law. $1$ mole of any gas at STP occupies a volume of $22.4{\text{L}}$.