Chapter 23, Problem 81A

Interpretation Introduction

Interpretation: To find the new pressure for a sample of nitrogen sealed in a container.

Concept Introduction: According to Gay-Lussac’s law, the pressure of a given mass of a gas varies directly with the absolute temperature of the gas when the volume is kept constant. Mathematically it is written as

  $\text{P}\propto \text{T}$ when the volume is constant.

  $\text{P=kT}$ where k = constant

  $\text{P/T=k}$ mathematical form

Answer to Problem 81A

115 kPa

Explanation of Solution

Given the pressure ${\text{P}}_{\text{1}}=1.00\times {10}^2\text{kPa}$

The initial temperature ${\text{T}}_{\text{1}}={25.0}^{\text{Ο}}\text{C}$ . On converting this temperature to kelvin, the temperature will be 298.15 K

The final temperature ${\text{T}}_{\text{2}}\text{=68}{\text{.7}}^{\text{Ο}}\text{C}$ . On converting this temperature to kelvin, the temperature will be 341.85 K

Now, the final pressure has to be determined.

  $\begin{array}{l}{\text{P}}_{\text{1}}{\text{/T}}_{\text{1}}{\text{=P}}_{\text{2}}{\text{/T}}_{\text{2}}\\ \text{1}{\text{.00×10}}^{\text{2}}\text{kPa/298}{\text{.15K=P}}_{\text{2}}\text{/341}\text{.7K}\\ \text{100kPa/298}{\text{.15K=P}}_{\text{2}}\text{/341}\text{.7K}\\ \text{100kPa}\times \text{341}\text{.7K/298}{\text{.15K=P}}_2\end{array}$

  ${\text{P}}_{\text{2}}\text{=114}\text{.606kPa}\approx \text{115kPa}$

Therefore, the new pressure is equal to $\text{115kPa}$ .

Conclusion

This law clearly explains that the volumes of the reacting gases are small whole numbers when measured at the same temperature and pressure. In this case, the volume is kept constant while the pressure and the temperature will vary.