Heat and Thermodynamics 6 Question 33
38. 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: 38. (a) $\uparrow _A^{p}{ }^{B}{ } _C V$
(b) $113 L, 0.44 \times 10^{5} N / m^{2}$ (c) $12459 J$
Solution:
- 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 $\left(p _1\right)$
$$
- \text { Pressure due to argon }\left(p _2\right) $$
or $p=p _1+p _2=\frac{n _1 R T}{V}+\frac{n _2 R T}{V}=\left(n _1+n _2\right) \frac{R T}{V}$
Substituting the values
$$
