Question

A closed container of volume $0.02m^3$ contains a mixture of neon and argon gases, at a temperature of ${27}^o$C and pressure of $1\times {10}^5$ $N/m^2$. The total mass of the mixture is $28$ gm. If the gram molecular weights of neon and argon are $20$ and $40$ respectively, find the masses of the individual gases in container, assuming them to be ideal.(Universal gas constant R$=8.314$ J/mol. K)

• A. $m_1=4.074$gm, $m_2=33.926$gm.
• B. $m_1=4.074$gm, $m_2=23.926$gm.
• C. $m_1=5.074$gm, $m_2=23.926$gm.
• D. $m_1=4.074$gm, $m_2=63.926$gm.

Answer 1

Answer: (B)

$m_1=4.074$gm, $m_2=23.926$gm.

$m_{Ne}+m_{Ar}=28gm$

$PV=(n_{Ne}+n_{Ar})RT$

$\frac{PV}{RT}=\frac{m_{Ne}}{20}+\frac{m_{Ar}}{40}=\frac{{10}^5\times 0.02}{8.314\times 300}$

$2m_{Ne}+m_{Ar}=32.0744$

$m_{Ne}=4.0744$

$m_{Ar}=23.926$