Chapter 10, Problem 105IL

You have a gas, one of the three known phosphorus-fluorine compounds (PF₃, PF₃, and P₂F₄). To find out which, you have decided to measure its molar mass.

1. (a) First, yon determine that the density of the gas is 5.60 g/L at a pressure of 0.971 atm and a temperature of 18.2 °C. Calculate the molar mass and identify the compound.
2. (b) To check the results from part (a), you decide to measure the molar mass based on the relative rales of effusion of the unknown gas and CO₂. You find that CO₂ effuses at a rate of 0.050 mol/min, whereas the unknown phosphorus fluoride effuses at a rate of 0.028 mol/min. Calculate the molar mass of the unknown gas based on these results.

Interpretation Introduction

Interpretation:

 From the given set of gases the gas that matches with the molar mass obtained from the calculation using given set of conditions should be determined.

Concept introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\\ \text{where,}\\ \text{n = moles of gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties. At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

The root mean square velocity $\mu$ is defined as the measure of velocity of particle in gas. It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

                                                        ${\mu }_{rms}={\left(\frac{3RT}{M}\right)}^{1/2}$ (1)

  $\begin{array}{l}\left(\text{gas constant}\right)R=\frac{8.314J}{Kmol}\\ M=Molarmass\end{array}$

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Average Kinetic energy: The kinetic energy for the gas is directly proportional to the kelvin temperature. The kinetic energy is equal to half of the multiplied value obtained by multiplication of mass of gas with $\text{rms}$ velocity of the gas.

Answer to Problem 105IL

The molar mass for the gas is found to be $\text{138g/mol}$ and hence the gas is found to be ${\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Density, D}\text{=5}\text{.60 g/L}\\ \text{Pressure, P}\text{=}\text{0}\text{.971 atm}\\ \text{Temperature,}\text{T = 18}{\text{.2}}^{\text{o}}\text{C = 273}\text{.15+18}\text{.2 = 291}\text{.35K}\\ \end{array}$

Using ideal gas equation the molar mass for the unknown gas is determined as follows,

  $\begin{array}{l}\text{PV= nRT}\\ \text{P = }\frac{\text{nRT}}{\text{V}}\\ \text{P = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\text{×}\frac{\text{RT}}{\text{V}}\\ \text{Molar}\text{mass = }\frac{\text{Mass}}{\text{V}}\text{×}\frac{\text{RT}}{\text{P}}\\ \text{=5}\text{.60 g/L×}\frac{\text{0}\text{.0821×291}\text{.35}}{\text{0}\text{.971}}\left[\text{since }\frac{\text{Mass}}{\text{V}}\text{ = D}\right]\\ \text{=137}\text{.95g/mol}\\ \text{=138 g/mol}\end{array}$

From the above calculation it is clear that the molar mass for the given unknown gas is found to be $\text{138 g/mol}$ which actually equals to the molar mass of ${\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$ from the given set of gases.

Therefore, the unknown gas is found to ${\text{P}}_{\text{2}}{\text{F}}_{\text{4}}$.

Interpretation Introduction

Interpretation:

 From the given set of gases the gas that matches with the molar mass obtained from the calculation using given set of conditions should be determined.

Concept introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\\ \text{where,}\\ \text{n = moles of gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties. At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

The root mean square velocity $\mu$ is defined as the measure of velocity of particle in gas. It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

                                                        ${\mu }_{rms}={\left(\frac{3RT}{M}\right)}^{1/2}$ (1)

  $\begin{array}{l}\left(\text{gas constant}\right)R=\frac{8.314J}{Kmol}\\ M=Molarmass\end{array}$

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Average Kinetic energy: The kinetic energy for the gas is directly proportional to the kelvin temperature. The kinetic energy is equal to half of the multiplied value obtained by multiplication of mass of gas with $\text{rms}$ velocity of the gas.

Answer to Problem 105IL

The molar mass for the given gas is $\text{140g/mol}$.

Explanation of Solution

Using the effusion rate of the unknown gas compared with the known ${\text{ CO}}_{\text{2}}$ gas the above calculation is verified as follows,

  $\begin{array}{l}\frac{{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}}\text{=}\sqrt{\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}}\\ \text{where,}{\text{r}}_{\text{1}}\text{=}{\text{Effusion rate for CO}}_{\text{2}}\\ {\text{r}}_{\text{2}}\text{=}\text{Effusion rate for unknown gas}\\ {\text{M}}_{\text{2}}\text{=}\text{Molar mass of unknown gas}\\ {\text{M}}_{\text{1}}\text{=}{\text{Molar mass for CO}}_{\text{2}}\\ \frac{\text{0}\text{.050 mol/min}}{\text{0}\text{.028 mol/min}}\text{= }{\left(\frac{{\text{M}}_{\text{2}}}{\text{44}\text{.01}}\right)}^{\text{1/2}}\\ {\text{M}}_{\text{2}}\text{=}\text{140g/mol}\\ \end{array}$

The above calculation shows that the molar mass for unknown gas is found to be $\text{140g/mol}$. Therefore, it verifies the answer in the part a since the molar mass obtained from both the calculation matches.