Chapter U3, Problem 14STP

Interpretation Introduction

Interpretation:

The final pressure of a 8.5 L scuba tank as it is cooled from an initial temperature of 26 ^oC and pressure of 150 atm to temperature of 15 ^oC needs to be calculated.

Concept introduction:

The relation between Temperature (T) and pressure (P) of a gas under constant volume is represented through Gay-Lussac's Law.

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

Answer to Problem 14STP

(C) 140 atm

Explanation of Solution

Given:

Initial temperature, T₁=26° C

Final temperature, T₂ = 15°C

Initial pressure, P₁ = 150 atm

Calculation:

Let P₂ represent the final pressure of the tank:

Based on equation (1):

$\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

$\begin{array}{l}{\text{P}}_2=\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\times {\text{T}}_2\\ \text{=}\frac{\text{150 atm}}{\text{(26+273)K}}\times (15+273)\text{K = }144\text{ atm}\end{array}$

This is close to option (c).

Conclusion

The pressure of the tank at a final temperature of 15 ^oC is around 140 atm.