Question

Bacteria and viruses are inactivated by temperatures above $135°C$. An autoclave contains steam at $1.00atm$and $100°C$. What is the pressure, in atmospheres, when the temperature of the steam in the autoclave reaches $135°C$if $Vandn$do not change?

Short answer

The final pressure of steam autoclave is $1.1atm$.

Step-by-step solution

Given Information  

We need to find the final temperature of the steam autoclave.

Simplify

Considering the Gay- Lussac's Law that if a constant volume and amount of gas are maintained, the pressure will be increased. In the temperature-pressure relationship which is known as Gay-Lussac's law, the pressure of a gas decreases, and a decrease in temperature decreases the pressure of the gas till the volume of the gas doesn't change. i.e.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}...\left(1\right)$

Calculation

Here we have the quantities available, that is

the initial temperature $T_1=100°C=\left(100+273\right)K=373K$

initial pressure$P_1=1.00atm$

the final temperature$T_2=135°C=\left(135+273\right)K=408K$

Now, calculating the final pressure $P_2$from the equation $\left(1\right)$

$$\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ {P}_2=\frac{P_1\times T_2}{T_1} \end{gathered}$$

Putting the values:

$$\begin{gathered} P_2=\frac{1.0atm\times 408K}{373K} \\ {P}_2=\frac{408atm}{373} \\ =1.1atm \end{gathered}$$