Question

The gauge pressure in a helium gas cylinder is initially 32 atm. After many balloons have been blown up, the gauge pressure has decreased to 5 atm. What fraction of the original gas remains in the cylinder?

Short answer

Finally, \(18\% \) of the original gas is present in the cylinder.

Step-by-step solution

Concepts

The pressure of a gas is equal to the gauge pressure and the pressure of the atmosphere.

For this problem, you have to use\(\frac{{{P_1}}}{{{n_1}}} = \frac{{{P_2}}}{{{n_2}}}\)equation.

Given data

The initial gauge pressure of the helium gas is\({P_1} = 32\;{\rm{atm}}\).

The final gauge pressure of the helium gas is\({P_2} = {\rm{5}}\;{\rm{atm}}\).

Let there be\({n_1}\)mole helium initially and\({n_2}\)mole helium finally in the cylinder.

The volume of the cylinder is constant.

Calculation

Now, from the ideal gas law, you get:

\(\begin{aligned}{c}PV &= nRT\\\frac{P}{n} &= \frac{{RT}}{V}\end{aligned}\)

The initial pressure of the gas is:

\(\begin{aligned}{c}{{P'}_1} &= 32\;{\rm{atm}} + 1\;{\rm{atm}}\\ &= 33\;{\rm{atm}}\end{aligned}\)

The final pressure of the gas is:

\(\begin{aligned}{c}{{P'}_2} &= 5\;{\rm{atm}} + 1\;{\rm{atm}}\\ &= 6\;{\rm{atm}}\end{aligned}\)

As the volume of the cylinder is constant,

\(\begin{aligned}{c}\frac{{{{P'}_1}}}{{{n_1}}} &= \frac{{{{P'}_2}}}{{{n_2}}}\\\frac{{{n_2}}}{{{n_1}}} &= \frac{{{{P'}_2}}}{{{{P'}_1}}}\\\frac{{{n_2}}}{{{n_1}}} &= \frac{{6\;{\rm{atm}}}}{{33\;{\rm{atm}}}}\\\frac{{{n_2}}}{{{n_1}}} &= \frac{6}{{33}}\end{aligned}\)

On further calculation, you get:

\(\begin{aligned}{c}\frac{{{n_2}}}{{{n_1}}} \times 100\% &= \frac{6}{{33}} \times 100\% \\ &= 18.2\% \\ \approx 18\% \end{aligned}\)

Hence, finally, \(18\% \) of the original gas is present in the cylinder.