Chapter U3, Problem 9STP

Interpretation Introduction

Interpretation:

The new volume of the balloon as it expands from an initial volume of 1.35 L and temperature 22 ^oC to a final temperature of 32 ^oC needs to be calculated.

Concept introduction:

The relation between volume (V) and temperature (T) of a gas under constant pressure and amount of the gas is represented through Charles Law.

As per Charles law, volume is directly related to temperature:

$\begin{array}{l}\text{V α T i}\text{.e}\text{. V = k}\times \text{T}\\ \text{where 'k' is the proportionality constant}\\ (\text{or})\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{V}}_2}{{\text{T}}_2}-----(1)\\ {\text{where V}}_{\text{1}}{\text{ and V}}_{\text{2}}\text{ are the initial and final volume}\\ {\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}\text{ are the initial and final temperature}\end{array}$

Answer to Problem 9STP

(C) 1.4 L

Explanation of Solution

Given:

Initial volume of balloon, V₁ = 1.35 L

Initial temperature, T₁ = 22°C

Final temperature, T₂ = 32°C

Calculation:

Based on equation (1):

$\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{V}}_2}{{\text{T}}_2}$

$\begin{array}{l}{\text{V}}_2=\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\times {\text{T}}_2\\ \text{=}\frac{\text{1}\text{.35 L}}{\text{(22+273)K}}\times (32+273)\text{K = 1}\text{.39 L}\simeq \text{1}\text{.4 L}\end{array}$

Conclusion

The volume of balloon at a final temperature of 32 ^oC is 1.4 L.