Question

Rahul forgot to add the reaction mixture to the round-bottomed flask at 27 degrees celsius but instead, he placed the flask on the flame. After a lapse of time, he realized his mistake and using a pyrometer he found the temperature of the flask was 477 degrees Celcius. What fraction of air would have been expelled out?

Answer 1

Step 1: Given data

${\mathrm{T}}_1={27}^{\mathrm{o}}\mathrm{C}=300\mathrm{K};$

${\mathrm{T}}_2={477}^{\mathrm{o}}\mathrm{C}=750\mathrm{K}$

Step: 2 Introducing the formula :

Charles law states that "When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion".

According to Charles law equation: V ∝ T

The mathematical expression of Charles law can be expressed as follows for comparing the same material under two different sets of conditions:

$$\frac{V_1}{V_2}=\frac{T_1}{T_2}$$

$\Rightarrow \frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where V₁ is the initial volume of the system.

V₂ is the final volume of the system.

T₁ is the initial temperature of the system in Kelvin.

T₂ is the final temperature of the system in Kelvin.

Step: 3 Putting the values in the above-mentioned equation

$$\begin{gathered} \frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{300}{750}=0.4 \end{gathered}$$

The fraction of air expelled out

= $1-0.4=0.6$
Therefore, the volume of air that was expelled out is $0.6$