Question

A space capsule is filled with neon gas at \[1.00\]atm and \[290K\]. The gas effuses through a pinhole into the outer space at such a rate that the pressure drops by \[0.3\]torr/sec
A.If the capsule were filled with ammonia at the same temperature and pressure, what would be the rate of pressure drop?
B.If the capsule were filled with \[30.0\% \] helium, 20.0% oxygen and 50.0 mol % nitrogen at a total pressure of 1.00 atm and a temp of 290 K, what would be the corresponding rate of pressure drop.

Answer 1

Hint: The given question is related to Graham’s law of effusion also known as Graham’s law of diffusion. Let us understand this law and solve the given problem with the formula given below. It is related to mass of particles and effusion of gas.

Formula used: \[\dfrac{{r{}_1}}{{r{}_2}} = \sqrt {\dfrac{{M{}_1}}{{M{}_2}}} \] where,
\[r{}_1\]= rate of effusion of given gas 1
\[r{}_2\]= rate of effusion of given gas 2
\[M{}_1\]= Molar mass of given gas 1
\[M{}_2\]= Molar mass of given gas 1

Complete step by step answer:
The rate of pressure drop is directly related to rate of effusion.
$\dfrac{{r{}_{Ne}}}{{r{}_{NH{}_3}}} = \sqrt {\dfrac{{M{}_{NH{}_3}}}{{M{}_{Ne}}}} = \sqrt {\dfrac{{17.0}}{{20}}} = 0.92 \\
  r{}_{NH{}_3} = \dfrac{{r{}_{Ne}}}{{0.92}} = \dfrac{{0.30}}{{0.92}} = 0.326 \\
$
Hence, \[0.326\]torr/sec is the rate of effusion.
B.The average molecular mass of the gaseous mixture is
\[0.3 \times 4 + 0.2 \times 32 + 0.5 \times 28 = 21.6\]
The rate of drop of pressure
$= \dfrac{{20}}{{21.6}} \times 0.30 \\
= 0.29 \\
$
\[0.29\]torr/sec is rate of drop of pressure

So, the correct answer is Option E.

Note: Graham’s law of effusion is inversely proportional to the square root of its molecular weight. This law is the most accurate for the effusion of molecules involving the movement of one gas at a time through a tiny pin hole.