Chapter U3, Problem 16STP

Interpretation Introduction

Interpretation:

The final volume of a 7500 L helium gas balloon as it cools from an initial temperature of 24° C and pressure of 1 atm to a final temperature of -15° C and 0.001 atm pressure, needs to be calculated.

Concept introduction:

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

As per Boyles law, pressure is inversely related to volume:

$\begin{array}{l}\text{P α }\frac{\text{1}}{\text{V}}\text{ i}\text{.e}\text{. P = k}\times \frac{\text{1}}{\text{T}}\\ \text{where k is a proportionality constant}\\ {\text{(or) P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = P}}_{\text{2}}{\text{V}}_{\text{2}}\text{ -----(2)}\end{array}$

As per Gay-Lussac’s law, pressure is directly related to temperature: $\begin{array}{l}\text{P α T i}\text{.e}\text{. P = k}\times \text{T}\\ \text{where 'k' is the proportionality constant}\\ (\text{or})\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{2}}}-----(3)\\ {\text{where P}}_{\text{1}}{\text{ and P}}_{\text{2}}\text{ are the initial and final pressures}\\ {\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}\text{ are the initial and final temperature}\end{array}$

Equations (1), (2) and (3) give the combined gas law according to which:

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{--------(4)}$

Answer to Problem 16STP

(B) 6.5x10⁶ L

Explanation of Solution

Given:

Initial temperature, T₁=24° C

Final temperature, T₂ = - 15°C

Initial pressure, P₁ = 1 atm

Final pressure, P₂ = 0.001 atm

Initial volume, V₁ = 7500 L

Calculation:

Let V₂ represent the final volume of the He gas balloon

Based on equation (4):

$\begin{array}{l}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\\ {\text{V}}_2=\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\times \frac{{\text{T}}_2}{{\text{P}}_{\text{2}}}\\ \text{=}\frac{\text{1 atm}\times \text{7500 L}}{\text{(24+273)K}}\times \frac{(-15+273)\text{K }}{0.001\text{ atm}}\text{= }6.5\times {10}^6\text{ L}\end{array}$

Conclusion

The final volume of the He gas balloon is 6.5x10⁶ L.