Chapter 13, Problem 36P

To determine

To Find: The massof helium needed.

Answer to Problem 36P

The mass of helium needed is $2.68\text{kg}$

Explanation of Solution

Given:

Gauge pressure of the O₂= $\text{8}.\text{7}0\text{atm}$

Gauge pressure of the He= $\text{7}.00\text{atm}$

Formula used:

Ideal gas law Equation

  $\begin{array}{l}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\\ \frac{\mathrm{P}}{\mathrm{n}}=\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\\ \frac{{\mathrm{P}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{P}}_2}{{\mathrm{n}}_2}\\ {\mathrm{n}}_2={\mathrm{n}}_1\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\end{array}$

Here, ${\mathrm{n}}_1$ is the number of moles for gas 1,

  ${\mathrm{P}}_1$ is the gauge pressure for gas 1,

  ${\mathrm{P}}_2$ is the gauge pressure for gas 2.

Calculation:

Oxygen and helium assume to be ideal gases. Means volume and temperature constant for both of them.

From ideal gas law

  $\begin{array}{c}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\end{array}$

  $\begin{array}{c}\frac{\mathrm{P}}{\mathrm{n}}=\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\end{array}$

  $\begin{array}{c}\frac{{\mathrm{P}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{P}}_2}{{\mathrm{n}}_2}\end{array}$

  $\begin{array}{c}{\mathrm{n}}_2={\mathrm{n}}_1\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\end{array}$

  $\begin{array}{c}=(26.0\text{kg}{\text{O}}_{\text{2}})\left(\frac{1\text{mole}{\text{O}}_{\text{2}}}{32\times {10}^{-3}\text{kg}}\right)\frac{\left(8.00\text{atm}\right)}{\left(9.70\text{atm}\right)}=6.70\times {10}^2\text{moles}\end{array}$

  $\begin{array}{c}=(6.70\times {10}^2\text{moles})\left(\frac{4.0\times {10}^{-3}\text{kg}}{1\text{mole}\text{He}}\right)=2.68\text{kg}\text{He}\end{array}$

Conclusion:

The mass needed is 2.68 kg.