Chapter 5, Problem 5.33QP

(a)

Interpretation Introduction

Interpretation:

Considering the setup given, the container with the greater density should be identified. Should be measured.

Concept Introduction:

Using the ideal gas law to find molar mass and density of a gas.

$\begin{array}{l}\text{PV = nRT}\\ \text{n = }\frac{\text{PV}}{\text{RT}}\\ \text{n =}\frac{\text{mass}}{\text{molar}\text{mass}}\text{ = }\frac{\text{PV}}{\text{RT}}\\ \frac{\text{mass}}{\text{V}}\text{= }\frac{\text{P}\text{.molarmass}}{\text{RT}}\left(\text{Density=}\frac{\text{mass}}{\text{volume}}\right)\\ \text{D =}\frac{\text{P}\text{.}\text{molarmass}}{\text{RT}}\end{array}$

Answer to Problem 5.33QP

The container with ${\text{O}}_{\text{2}}$ gas has higher density.

Explanation of Solution

From the ideal gas equation and density formula,

Both the density and pressure are related.

$\text{D =}\frac{\text{P}\text{.}\text{molarmass}}{\text{RT}}$

From the above equation we can say that, pressure and density are directly proportional to each other. So the molecules with the highest molecular mass will have higher density i.e., O₂ gas with $\text{molar mass}\text{=}\text{32}\text{.00}\text{g/mol}$ where as ${\text{H}}_{\text{2}}\text{ gas with molar mass}\text{=}\text{2}\text{.016}\text{g/mol}$

Conclusion

The container with the greater density was identified.

(b)

Interpretation Introduction

Interpretation:

Considering the setup given, the container with the greater RMS speed should be identified.

Concept Introduction:

Root mean square value:

${\text{μ}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{m}}}}$

Where,

$\begin{array}{l}{\text{μ}}_{\text{rms}}\text{=}\text{root}\text{mean}\text{square}\text{value}\\ \text{R}\text{= ideal}\text{gas}\text{constant = 8}\text{.314}\frac{\text{joule}}{\text{mole}\text{.Kelvin}}\\ \text{K}\text{=}\text{absolute}\text{temperature}\text{in}\text{kelvin}\\ \text{M}\text{=}\text{molar}\text{mass}\text{of}\text{the}\text{molecule}\text{in}\text{kilograms}\end{array}$

Answer to Problem 5.33QP

The ${\text{H}}_{\text{2}}$ gas molecule has greater RMS speed.

Explanation of Solution

From the root mean square value we can say that the molecule with less molar mass $\left({\text{H}}_{\text{2}}\text{=2}\text{.016}\text{g/mol}\right)$ will have higher velocity than the O₂ gas with $\text{molar mass}\text{=}\text{32}\text{.00}\text{g/mol}$.

Conclusion

The change in mole fraction of O₂ at lower temperature in the flask was explained.

(c)

Interpretation Introduction

Interpretation:

Considering the setup given, the container with the greater number of atoms should be identified.

Concept Introduction:

Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

$\begin{array}{l}\text{Vα n (at}\text{a}\text{constant}\text{pressure)}\\ \text{or}\\ \frac{\text{V}}{\text{n}}\text{= k}\end{array}$

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Answer to Problem 5.33QP

The amount of atoms present in the containers are same.

Explanation of Solution

Both the containers has 1.0 moles of gas, so from the Avogadro law we can say that they have equal number of atoms irrespective of their molar mass.

Conclusion

Considering the setup given, the container with the greater number of atoms was identified.

(d)

Interpretation Introduction

Interpretation:

Considering the setup given, the change in the pressure of the container when the closed valve is opened should be identified.

Concept Introduction:

Gas pressure:

Pressure or Stress is the force applied perpendicular to the surface of an object per unit area.

SI derived unit of pressure is Pascal (Pa).

Answer to Problem 5.33QP

The pressure in both the containers remains same.

Explanation of Solution

The pressure in the container remains same. Since both the container starts with the same pressure from their ends. so the total pressure remains same.

Conclusion

Considering the setup given, the change in the pressure of the container when the closed valve is opened was identified.

(e)

Interpretation Introduction

Interpretation:

Considering the given setup, the H₂ fraction of pressure when $\text{2}\text{.0}\text{mol}\text{of}\text{Ar}\text{gas}$ is allowed through opened valve of the container should be identified

Concept Introduction:

Gas pressure:

Pressure or Stress is the force applied perpendicular to the surface of an object per unit area.

SI derived unit of pressure is Pascal (Pa).

Answer to Problem 5.33QP

The fraction of total pressure due to H₂ gas can be ¼

Explanation of Solution

When the total amount of the gas contains in a container changes from 2.0 mol to 4.0 mol. the fraction of gas present in the container will also change, i.e, when two equal amount gases were present they are 50%  each in the container. When 2.0 mol of an argon gas introduced through the valves, the percentage of H₂ gas now changes 25 % that means ¼ of the container.

Total volume before the addition of Argon gas = 2.0 mol

Percentage of H₂ gas = 50 %( ½ fraction of the container)

Total volume after the addition of Argon gas = 4.0 mol

Percentage of H₂ gas = 25 %( ¼ fraction of the container)

Conclusion

Considering the given setup, the H₂ fraction of pressure when $\text{2}\text{.0}\text{mol}\text{of}\text{Ar}\text{gas}$ is allowed through opened valve of the container was explained.