Chapter 16, Problem 120A

Interpretation Introduction

Interpretation: The pressure of nitrogen gas is to be calculated when the temperature changes from $25°\text{C}$ to $45°\text{C}$ and is at a pressure of $101.3\text{kPa}$ .

Concept introduction: Gay-Lussac's law is one of the important gas laws which tells about the relationship between the temperature and pressure of gases. There is a direct relationship exists between gases. That is when temperature increases the pressure of gases will also increase.

It can be expressed in a mathematical form as follows:

  $P\propto T$

  $\frac{P}{T}=K$

Here $P$ is pressure, $T$ is temperature and $K$ is proportionality constant.

The formula can also be expressed as,

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$   ...... (1)

Here, $P_1$ is the initial pressure, $P_2$ is the final pressure, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Answer to Problem 120A

The new pressure of nitrogen gas when the temperature changes from $25°\text{C}$ to $45°\text{C}$ is $108.09\text{kPa}$ .

Explanation of Solution

The initial pressure is $101.3\text{kPa}$ .

The initial temperature is $25°\text{C}$ .

Final temperature is $45°\text{C}$ .

Convert temperate from degree Celsius to Kelvin.

  $\begin{array}{c}{T}_1=25°\text{C}\\ =25+273\\ =298\text{K}\end{array}$

  $\begin{array}{c}{T}_2=45°\text{C}\\ =45+273\\ =318\text{K}\end{array}$

Substitute $318\text{K}$ for $T_2$ , $298\text{K}$ for $T_1$ and $318\text{K}$ for $P_1$ in equation (1).

  $\begin{array}{c}{P}_2=101.3\times \frac{318}{298}\\ =108.09\text{kPa}\end{array}$

Conclusion

The new pressure of nitrogen gas when the temperature changes from $25°\text{C}$ to $45°\text{C}$ is $108.09\text{kPa}$ .