Chapter 13, Problem 36A

Interpretation Introduction

Interpretation: -Partial pressure is the pressure applied by a gas alone in a container.

Concept Introduction: - The total pressure of a mixture of gases which do not react chemically is equal to the sum of the partial pressure of the component gases.

  $\mathrm{P}=P_1+P_2+P_3\mathrm{.......}+P_n$

Answer to Problem 36A

Partial pressure of ${\mathrm{O}}_2=732mmHg$ and $0.0197moles$ of Oxygen are used.

Explanation of Solution

For calculating partial pressure of component gas and the total pressure, the Dalton’s law partial pressure is applied.

According to this law

  $\begin{array}{l}P=P_1+P_2\\ P_{total}=P_{O_2}+P_{H_{2}o}\\ 755mmHg=P_{O_2}+23mmHg\end{array}$

  $\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}=\text{755mmHg - }23\mathrm{mmHg}\\ {\text{P}}_{{\text{O}}_{\text{2}}}=\text{732mmHg}\end{array}$

Thus, Partial pressure of Oxygen ( ${\mathrm{O}}_2$ ) is 732mmHg

no. of moles Oxygen =?

  $\mathrm{PV}=nRT$

  $\begin{array}{c}\text{P = 732mmHg}\\ \text{1atm = 760mmHg}\\ \frac{732\mathrm{mmHg}}{760\mathrm{mmHg}}\end{array}$

  $=\frac{732}{760}=0.963\mathrm{atm}$

P = 0.963atm

  $\begin{array}{l}T=24^{0}C+273\\ =297K\end{array}$

  $\begin{array}{l}V=500ml\\ 1l=1000ml\\ \frac{500ml}{1000ml}=\frac{5}{10}=0.5L\end{array}$

V = 0.5L

R = 0.0821

  $\mathrm{PV}=nRT$

  $\begin{array}{l}n=\frac{PV}{RT}\\ n=\frac{0.963atm\times 0.5L}{0.0821\frac{atm}{kmol}\times 297K}\\ n=\frac{0.963\times 0.5}{0.0821\times 297}=\frac{0.48155}{24.3837}\\ n=0.0197moles\end{array}$

Number of moles of Oxygen ( $O_2$ ) is 0.0197 moles.

Conclusion

Partial pressure of ${\mathrm{O}}_2=732mmHg$ and $0.0197moles$ of Oxygen are used.