Chapter 13, Problem 59A

Interpretation Introduction

Interpretation:

By changing the temperature, the volume of the gas is to be determined.

Concept introduction:

According to Charles’s law, the volume of the gas directly varies with its temperature at constant pressure. Its mathematical formulation is:

  $\begin{array}{l}\text{V}\propto \text{T}\\ \text{V}\text{=}\text{kT}\\ \text{Where,}\text{k}\text{is}\text{proportionally}\text{constant}\end{array}$

V = volume of the gas

T = temperature of the gas

Answer to Problem 59A

The volume of the given gas sample is $\text{0}\text{.922}\text{L}$ .

Explanation of Solution

Given information:

Initial temperature, (T₁) = $\text{22°C}$

Initial temperature in K= $\text{22+273}\text{=}\text{295K}$

Final temperature (T₂) = $\text{100°C}$

Or, in Kelvin = $\text{100+273 =373K}$

Initial volume (V₁) = $\text{729mL}$

Or, in Liter = $\text{0}\text{.729}\text{L}$

Calculation:

According to Charles’s law,

  $\begin{array}{l}\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\\ {\text{V}}_{\text{1}}{\text{T}}_{\text{2}}\text{=}{\text{V}}_{\text{2}}{\text{T}}_{\text{1}}\\ {\text{V}}_{\text{2}}\text{=}\frac{{\text{V}}_{\text{1}}\text{T}}{{\text{T}}_{\text{1}}}\\ \text{=}\frac{\text{373×0}\text{.729}}{\text{295}}\\ \text{=}\text{0}\text{.922}\text{L}\end{array}$

Conclusion

The volume of the gas is $\text{0}\text{.922}\text{L}$ .