Q. 8.28 - An Introduction to General, Organic, and Biological Chemistry

Question

A gas has a volume of$4.00 \mathrm{L}$ at $0 °\mathrm{C}.$ What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if $P$and $n$ do not change?

a. $1.50 \mathrm{L}$

 b . $1200 \mathrm{mL}$ 

c. $10.0 \mathrm{L}$ 

d. $50.0 \mathrm{mL}$

Step-by-Step Solution

Answer

  The final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, 

a.$-170°C$

b. $-191°C$

c. $409°C$

d.$-269.5°C$

Part(a) Step 1: Given information

We need to find that final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if Pressure and amount do not change. 

Part(a) Step 2: Explanation

Here we will use charles law becuase pressure and amount of gas do not change .

Here $T_2$ Is Required temperature ;

Also , ${\mathrm{T}}_2=0+273 \mathrm{K}=273 \mathrm{K}$

${\mathrm{T}}_2=\frac{{\mathrm{T}}_1\times {\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{273 \mathrm{K}\times 1.50}{4}=102 \mathrm{K}=102-273=-170°\mathrm{C}$

Part(b) Step 1: Given information

We need to find that final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if Pressure and amount do not change 

Part(b) Step 2: Explanation

Here we will use Charles's law because pressure and amount of gas do not change.

Also,${\mathrm{T}}_2=0+273 \mathrm{K}=273 \mathrm{K}$

$$\begin{gathered} {\mathrm{T}}_2=\frac{{\mathrm{T}}_1\times {\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{273 \mathrm{K}\times 1.20}{4}=81.9 \mathrm{K} \\ =81.9-273=-191°\mathrm{C} \end{gathered}$$

Part(c) Step 1: Given information

We need to find that final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if Pressure and amount do not change.

Part(c) Step 2: Explanation

Here we will use Charles's law because pressure and amount of gas do not change.

Also, ${\mathrm{T}}_2=0+273 \mathrm{K}=273 \mathrm{K}$

$$\begin{gathered} {\mathrm{T}}_2=\frac{{\mathrm{T}}_1\times {\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{273 \mathrm{K}\times 10.0}{4}=682.5 \mathrm{K} \\ =682.5-273=409°\mathrm{C} \end{gathered}$$

Part(d) Step 1: Given information

We need to find that final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if Pressure and amount do not change. 

Part(d) Step 2: Explanation

Here we will use Charles's law because pressure and amount of gas do not change.

Also, ${\mathrm{T}}_2=0+273 \mathrm{K}=273 \mathrm{K}$

$$\begin{gathered} {\mathrm{T}}_2=\frac{{\mathrm{T}}_1\times {\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{273 \mathrm{K}\times 0.050}{4}=3.14 \mathrm{K} \\ =3.14-273.15=-269.5°\mathrm{C} \end{gathered}$$