Chapter 13, Problem 41A

Interpretation Introduction

Interpretation: -Volume of a mixture of a gas needs to be determined using the values given.

Concept Introduction: - For calculating the volume of mixture of gas using Ideal gas equation,

$PV=n\text{R}T$

For mixture of gases.

$\mathrm{PV}=(n_1+n_2)RT$

Where

P = Pressure of gas

V = volume of gas

T = temperature

R = gas constant

n = no. of moles of gas.

$n=(n_1+n_2)$

${\mathrm{n}}_1$ = no. of moles of gas − A

 = no. of moles of gas − B

Answer to Problem 41A

Volume of a mixture of gas is $357.5L$ .

Explanation of Solution

There is a relationship between volume of a gas and number of molecules which expressed by Avogadro’s law

$V\propto N$

For calculating volume, first calculate the number of moles.

$\text{Number of moles}=\frac{\text{mass}}{\text{molar mass}}$

Calculate number of moles of He as follows:

$\begin{array}{l}(n)=\frac{14.2g}{4.002g/mol}\\ He=3.54moles.\end{array}$

Similarly, number of moles of Hydrogen can be calculated as follows:

$\begin{array}{c}\mathrm{n}=\frac{21.6g}{2.016g/mol}\end{array}$

 $\begin{array}{c}=10.71moles\end{array}$

Now,

$\begin{array}{l}Total{n}_T=n_1+n_2\\ n_T=3.54+10.71mol\\ n_T=14.25mol\end{array}$

According to Ideal gas Equation,

$\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$ {at standard temperature and pressure}

Here,

V =?

P = 0.985atm

T = 28 + 273

 = 301K

$\mathrm{R}=0.0821\frac{atmL}{Kmol}$

Putting the values,

$\begin{array}{l}V\text{ = }\frac{\text{14}\text{.25 mol × 0}\text{.0821atm L × 301k}}{\text{0}\text{.985atm × k mol}}\\ V\text{ = }\frac{\text{14}\text{.25 × 0}\text{.0821L × 301}}{\text{0}\text{.985}}\\ V\text{ = }\frac{352.14}{0.985}\\ V\text{ = 357}\text{.5 L}\end{array}$

Volume of mixture of gas is $357.5L$ .

Conclusion

Volume of a mixture of gas is $357.5L$ .