Question

If a 2 litre flask of \[N_2\] at 20℃ and 70cm of P is connected with a 3 litre of another flask of ${O_2}$ at the same temperature and $100cm$ of P. What will be the final pressure after the glass have thoroughly mixed at the same temperature as before? Also, calculate the mole $\% $ of each gas in the resulting mixture. The volume of the stopcock can be neglected.

Answer 1

Hint: The first step would be to find the moles of both ${N_2}$ and ${O_2}$. Here the Rate would be a variable. After that we need to use those moles to find the final pressure or the total pressure. Mole fraction can be easily calculated by using the mole of one gas as numerator and total mole as denominator.

Complete step by step solution:
Given information to us is:
Volume of ${O_2}$ : $3l$
Temperature of ${O_2}$ : ${20^ \circ }C$
Pressure of ${O_2}$ : $100cm$ of P
 Volume of ${N_2}$ : $2l$
Temperature of ${N_2}$ : ${20^ \circ }C$
Pressure of ${N_2}$ : $70cm$
Step1: We will calculate the mole of ${N_2}$ first. Let moles of ${N_2}$ be ${n_1}$.
${n_1} = \dfrac{{PV}}{{RT}}$, here the $P$ is the pressure, $V$ is the volume, $R$ is the rate, $T$ is the temperature all related to the ${N_2}$ as given in question.
$\Rightarrow$ ${n_1} = \dfrac{{70 \times 2}}{{R \times 20}}$
Step2: We will calculate the mole of ${O_2}$. Let the moles of ${O_2}$be ${n_2}$.
${n_2} = \dfrac{{PV}}{{RT}}$, here the $P$ is the pressure, $V$ is the volume, $R$ is the rate, $T$ is the temperature all related to the ${O_2}$ as given in question.
$\Rightarrow$ ${n_2} = \dfrac{{100 \times 3}}{{R \times 20}}$
Step3: Now once we have the moles we can calculate the final pressure.
Let final pressure be ${P_f}$.
${P_f} = \dfrac{{\left( {{n_1} + {n_2}} \right)RT}}{V}$, Here ${P_f}$ is the final pressure, $V$ is the total volume, $R$ is the final rate, $T$ is the temperature.
$\Rightarrow$ ${P_f} = \dfrac{{\left( {\dfrac{{70 \times 2}}{{R \times 20}} + \dfrac{{100 \times 3}}{{R \times 20}}} \right) \times R \times 20}}{{3 + 2}}$
$\Rightarrow$ ${P_f} = 170cm$
Step4: Now, since we already found the moles we can find mole fraction easily.
Mole fraction of ${N_2} = \dfrac{{{n_1}}}{{{n_1} + {n_2}}}$
$\Rightarrow$ ${N_2} = \dfrac{{\dfrac{{140}}{{20R}}}}{{\dfrac{{300}}{{20R}} + \dfrac{{140}}{{20R}}}}$
$\Rightarrow$ ${N_2} = 0.35$
Step5: Mole fraction of ${O_2}$ will be $1 - {N_2}$.
${O_2} = 1 - 0.35 = 0.65$.
Hence the final Pressure is $170 cm$ and the mole fraction of ${O_2}$ is $0.65$ and of ${N_2}$ is $0.35$.

Note: When the gas molecules collide with the surfaces of the object, it causes the gas pressure. The force of collision is so small but a smaller surface area often experiences a large no of collisions in a short time. This results in the pressure by gases which we saw in the question.