Question

A closed container of volume 0.02 m³ contains a mixture of neon and argon gases at a temperature of 27°C and pressure of 1 x 10^5N/m^2. The total mass of the mixture is 28 g.If the gram molecular weights of neon and argon are 20 and 40 respectively, find the mass of the individual gases in the container assuming them to be ideal. Given : R = 8.314 J/mol-K.​

Answer 1

Answer-

Mass of neon is 4 gram.

Mass of argon is 24 gram.

Explaination-

$\large{\underline{\sf{Given}}} - \\ \\ \bold{ \sf{(i)} \: Volume \: of \: closed \: container = 0.02 {m}^{3}} \\ \\ \bold{\sf{(ii)} \: Gases \: present \: in \: the \: container = Neon \: and \: Argon} \\ \\ \bold{\sf{(iii)} \: Pressure \: of \: the \: container = 1 \times {10}^{5} \: N/ {m}^{2}} \\ \\ \bold{\sf{(iv)} \: Total \: mass \: of \: the \: mixture = 28g} \\ \\ \bold{\sf{(v)} \: Temperature = 27 \degree} \\ \\ \sf{(vi)} \: R = 8.314 \: J/mol-K$

To Find-

(a) Mass of neon

(b) Mass of argon

$\large{\underline{\sf{Let}}} \\ \\ \sf{'m' \: be \: the \: mass \: of \: neon \: \: then,} \\ \\ \sf{The \: mass \: of \: argon = (28 - m)g.} \\ \\ \bold{\underline{Total \: number \: of \: moles \: of \: the \: mixture}} \\ \\ \rightarrow \sf{ \mu = \frac{m}{20} + \frac{(28 - m)}{40}} \\ \\ \rightarrow \sf{ \mu = \frac{m \times 2}{20 \times 2} + \frac{28 - m}{40} } \\ \\ \rightarrow \sf{ \mu = \frac{2m + 28 - m}{40} } \\ \\ \rightarrow \red{\boxed{\sf{ \mu = \frac{28 + m}{40}}}.......(1)}$

$\large \sf{Now -} \\ \\ \rightarrow\sf{\mu = \frac{PV}{RT}} \\ \\ \rightarrow\sf{\mu = \frac{1 \times {10}^{5} \times 0.02 }{8.314 \times 300}} \\ \\ \rightarrow\sf{\mu = \frac{ {10}^{3} \times 0.02 }{8.314 \times 3} } \\ \\ \rightarrow\sf{\mu = \frac{ {10}^{3} \times 20 }{8314 \times 3} } \\ \\ \rightarrow\sf{\mu = \frac{ \cancel{20000}}{ \cancel{24942}} } \\ \\ \rightarrow \red{ \boxed{\sf{\mu =0.8(approx)}}.......(2)}$

$\large \underline{ \sf{By \: solving \: (1) \: and \: (2)}} \\ \\ \rightarrow \sf{ \frac{28 + m}{40} = 0.8} \\ \\ \rightarrow \sf{28 + m = 0.8 \times 40} \\ \\ \rightarrow \sf{28 + m = 32} \\ \\ \rightarrow \sf{m = 32 - 28} \\ \\ \rightarrow \red{\sf{\boxed{m = 4 \: gram}}}$

$\large{\sf{And-}} \\ \\ \sf{Mass \: of \: argon = (28 - 4)g} \\ \\ \sf{Mass \: of \: argon = 24g}$

Answer 2

$\underline\mathtt \red{Answer}$

$\sf:mNe+mAr=28gm</p><p></p><p>$

PV=(n Ne +n Ar )RT RTPV = 20m Ne + 40m Ar = 8.314×30010 5 ×0.02 2mNe +m Ar =32.0744 m Ne=4.0744mAr =23.926

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