(II) What is the pressure inside a 38.0-L container holding 105.0 kg of argon gas at 21.6°C?

The ideal gas law is \(PV = nRT\) for n mole of ideal gas.

For this problem, first, find the given data in the SI unit and then put it in the ideal gas equation.

The initial temperature is \(T = {21.6^ \circ }{\rm{C}} = 294.6\;{\rm{K}}\).

The volume of the container is \(V = 38.0\;{\rm{L}} = 38.0 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{3}}}\)

The mass of argon inside the container is \(M = 105.0\;{\rm{kg}}\).

The mass of onemole of argon gas is \(m = 39.95\;{\rm{g/mol}} = 39.95 \times {10^{ - 3}}\;{\rm{kg/mol}}\).

Letbe the pressure of the gas inside the container.

Now the number of moles of the argon gas is \(n = \frac{M}{m}\).

Then from the ideal gas equation,you get the following:

\(\begin{array}{c}PV = nRT\\PV = \frac{M}{m}RT\\P = \frac{{MRT}}{{mV}}\end{array}\)

Now, substituting the values into the above equation,you getthe following:

\(\begin{array}{c}P = \frac{{\left( {105.0\;{\rm{kg}}} \right) \times \left( {8.314\;{\rm{J/mol}} \cdot {\rm{K}}} \right) \times \left( {294.6\;{\rm{K}}} \right)}}{{\left( {39.95 \times {{10}^{ - 3}}\;{\rm{kg/mol}}} \right) \times \left( {38.0 \times {{10}^{ - 3}}\;{{\rm{m}}^{\rm{3}}}} \right)}}\\ = 1.69 \times {10^8}\;{\rm{Pa}}\end{array}\)

Hence, the pressure onthe inside of the container is \(1.69 \times {10^8}\;{\rm{Pa}}\).