Question

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

Short answer

The pressure onthe inside of the container is $1.69\times {10}^8\mathrm{P}\mathrm{a}$.

Step-by-step solution

Concepts 

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.

Given data 

The initial temperature is $T={21.6}^{\circ }\mathrm{C}=294.6\mathrm{K}$.

The volume of the container is $V=38.0\mathrm{L}=38.0\times {10}^{-3}{\mathrm{m}}^3$

The mass of argon inside the container is $M=105.0\mathrm{k}\mathrm{g}$.

The mass of onemole of argon gas is $m=39.95\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}=39.95\times {10}^{-3}\mathrm{k}\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$.

Letbe the pressure of the gas inside the container.

Calculation

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:

$\begin{array}{c}P=\frac{\left(105.0\mathrm{k}\mathrm{g}\right)\times \left(8.314\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\right)\times \left(294.6\mathrm{K}\right)}{\left(39.95\times {10}^{-3}\mathrm{k}\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}\right)\times \left(38.0\times {10}^{-3}{\mathrm{m}}^3\right)}\\ =1.69\times {10}^8\mathrm{P}\mathrm{a}\end{array}$

Hence, the pressure onthe inside of the container is $1.69\times {10}^8\mathrm{P}\mathrm{a}$.