Heat And Thermodynamics Ques 210
- A closed container of volume $0.02 m^{3}$ contains a mixture of neon and argon gases, at a temperature of $27^{\circ} C$ and pressure of $1 \times 10^{5} Nm^{-2}$. The total mass of the mixture is $28 g$. If the molar masses of neon and argon are 20 and $40 g mol^{-1}$ respectively, find the masses of the individual gases in the container assuming them to be ideal.
(Universal gas constant $R=8.314 J / mol-K$ ).
(1994, 6M)
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Answer:
Correct Answer: 210.Mass of neon $=4.074 \mathrm{~g}$, mass of argon $=23.926 \mathrm{~g}$
Solution:
Formula:
- Given, temperature of the mixture, $T=27^{\circ} C=300 K$
Let $m$ be the mass of the neon gas in the mixture. Then, mass of argon would be $(28-m)$
Number of gram moles of neon, $n _1=\frac{m}{20}$ Number of gram moles of argon, $n _2=\frac{(28-m)}{40}$
From Dalton’s law of partial pressures.
Total pressure of the mixture $(p)=$ Pressure due to neon $\left(p_1\right)$ + Pressure due to argon $\left(p_2\right)$
or $\quad p=p_1+p_2=\frac{n_1 R T}{V}+\frac{n_2 R T}{V}=\left(n_1+n_2\right) \frac{R T}{V}$
Substituting the values $$ 210.0 \times 10^5=\left(\frac{m}{20}+\frac{28-m}{40}\right) \frac{(8.314)(300)}{0.02} $$
Solving this equation, we get
$$210-m=23.926 \mathrm{~g} \Rightarrow m=4.074 \mathrm{~g} $$
Therefore, in the mixture, $4.074 \mathrm{~g}$ neon is present and the rest i.e. $23.926 \mathrm{~g}$ argon is present.
BETA
