Chapter 11.4, Problem 42PP

Interpretation Introduction

Interpretation:

The final pressure of steam in atmosphere should be calculatedwhen the autoclave reaches a final temperature of $135\text{C}$ , under conditions of constant volume and amount of steam. Given that the steam in the autoclave is at initial pressure of 1.00 atm and temperature of $100\text{C}$ .

Concept introduction:

• The relation between pressure (P) and temperature (T) of a gas under constant volume and amount of the gas is represented through Gay-Lussac’s Law.
• As per Gay-Lussac’s law, pressure is directly related to temperature:

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

• The standard units for pressure is atmosphere (atm) and temperature kelvin (K)
• Pressure conversions:

1 torr = 1 mmHg

1 atm = 760 torr = 760 mmHg

Temperature conversion:

  $\text{K =}\text{C}\text{+ 273 (or) }\text{C}\text{= K}-\text{273}$

Given:

Initial pressure P₁ = 1.00 atm

Initial temperature T₁ = $100\text{C}$ = (100 + 273) K = 373 K

Final temperature T₂ = $135\text{C}$ = (135 + 273) K = 408 K

Final pressure = P₂