Chapter 10, Problem 10.29QA

Interpretation Introduction

To:

1. Calculate the room-mean-square speeds methane $\left(C{H}_4\right)$, ethane ${(C}_2{H}_6)$, and propane $\left(C_3{H}_8\right)$ at $298 K$.
2. Plot the root-mean-square speeds of these gases as a function of their molar mass and use it to predict the root-mean-square speed of butane $\left(C_4{H}_{10}\right)$ and pentane $\left(C_5{H}_{12}\right)$.c) Determine the gas that effuses from a sample of a natural gas (major components: methane, ethane, propane and butane) the most rapidly.

Answer to Problem 10.29QA

Solution:

1. ${\mu }_{rms} of C{H}_4=681\frac{m}{s}, {\mu }_{rms} of C_2{H}_6=497\frac{m}{s}, {\mu }_{rms} of C_3{H}_8=411\frac{m}{s}$
2. ${\mu }_{rms} of C_4{H}_{10}=359\frac{m}{s}, {\mu }_{rms} of C_5{H}_{12}=298\frac{m}{s}$
3. Methane effuses most rapidly.

Explanation of Solution

1)Concept:

1. To calculate the root-mean-square speed $\left({\mu }_{rms}\right)$ of all the given gases, we are using the formula of it given below in terms of gas constant $\left(R\right)$, temperature $\left(T\right)$, and molar mass $\left(\mathcal{M}\right).$
2. For this part, we plot the graph of root-mean-square speed in m/s vs reciprocal of square root of molar mass. Using the equation of straight line and the value of reciprocal of square root of molar mass for butane and pentane, we can get their root-mean-square speed.
3. Graham’s law of effusion states that the effusion rate of a gas is inversely proportional to the square root of its molar mass.

Therefore,

$Effusion rate=k\sqrt{\frac{1}{molar mass}}$

where k is constant.

So, the higher the molar mass of gas, the slower will be the effusion. And the smaller the molar mass, the higher is the effusion. In all these major compounds of natural gas, methane has the smaller molar mass. Therefore, it can effuse rapidly.

From the root-mean-square speed values, we can say that the higher the root-mean-square speed, the more rapid is the effusion of gas.

2)Formula:

${\mu }_{rms}=\sqrt{\frac{3RT}{\mathcal{M}}}$

3)Given:

$T=298 K$

 ii) ${\mathcal{M}}_{C{H}_4}=16.01\frac{g}{mol}, {\mathcal{M}}_{C_2{H}_6}=30.07\frac{g}{mol}, {\mathcal{M}}_{C_3{H}_8}=44.1\frac{g}{mol}, {\mathcal{M}}_{C_4{H}_{10}}=58.12\frac{g}{mol}, {\mathcal{M}}_{C_5{H}_{12}}=72.15\frac{g}{mol}$       

iii) $R=8.314\frac{kg.m^2}{s^2.mol.K}$

4)Calculation:

a) The unit of the gas constant is $kg$. So, we need to convert the molar masses to $kg/mol$.

${\mathcal{M}}_{C{H}_4}=16.01\frac{g}{mol}\times \frac{1 kg}{1000 g}=0.01601\frac{kg}{mol}$

${\mathcal{M}}_{C_2{H}_6}=30.07\frac{g}{mol}\times \frac{1 kg}{1000 g}=0.03007\frac{kg}{mol}$

${\mathcal{M}}_{C_3{H}_8}=44.1\frac{g}{mol}\times \frac{1 kg}{1000 g}=0.0441\frac{kg}{mol}$

${\mu }_{rms}=\sqrt{\frac{3RT}{\mathcal{M}}}$

${\mu }_{rms} of C{H}_4=\sqrt{\frac{3\times 8.314\frac{kg.m^2}{s^2.mol.K}\times 298 K}{0.01601\frac{kg}{mol}}}= \sqrt{464254.6\frac{m^2}{s^2}}=681\frac{m}{s}$

${\mu }_{rms} of C_2{H}_6=\sqrt{\frac{3\times 8.314\frac{kg.m^2}{s^2.mol.K}\times 298 K}{0.03007\frac{kg}{mol}}}= \sqrt{247180.4\frac{m^2}{s^2}}=497\frac{m}{s}$

${\mu }_{rms} of C_3{H}_8=\sqrt{\frac{3\times 8.314\frac{kg.m^2}{s^2.mol.K}\times 298 K}{0.0441\frac{kg}{mol}}}= \sqrt{168542.3\frac{m^2}{s^2}}=411\frac{m}{s}$

b) ${\mathcal{M}}_{C_4{H}_{10}}=58.12\frac{g}{mol}\times \frac{1 kg}{1000 g}=0.05812\frac{kg}{mol} \Rightarrow \frac{1}{\sqrt{\mathcal{M}}}=4.15$

${\mathcal{M}}_{C_5{H}_{12}}=72.15\frac{g}{mol}\times \frac{1 kg}{1000 g}=0.07215\frac{kg}{mol} \frac{1}{\sqrt{\mathcal{M}}}=3.72$

The plot is as follows:

To determine the values of root-mean-square speed, we draw a line from the $1/\sqrt{\mathcal{M}}$ value for C₄H₁₀ on the x-axis and from its intersection with the plot, draw a line to the y-axis to get the corresponding $u_{rms}$ value. Similarly, draw a line from the $1/\sqrt{\mathcal{M}}$ value for C₅H₁₂ on the x-axis and from its intersection with the plot, draw a line to the y-axis to get the corresponding $u_{rms}$ value.

The values of $u_{rms}$ obtained from the plot are: 359 m/s for C₄H₁₀ and 298 m/s for C₅H₁₂.

Conclusion:

Using molar masses and given temperature, we got root-mean-square speed of natural gases.

Smaller the molar mass, fast it can effuse. Therefore, methane can effuse fast than other given gases.