The value of coefficient of volume expansion of glycerin is `5 xx 10^(-4) K^(-1)`. The fractional change in the density of glycerin for a rise of `40^(@)C` in its temperature is

Options

A. 0.015
B. 0.02
C. 0.025
D. 0.01

Answer

To solve the problem, we need to find the fractional change in the density of glycerin given its coefficient of volume expansion and a temperature change. Let's go through the solution step by step.

### Step 1: Understand the relationship between volume expansion and density
The coefficient of volume expansion (\(\alpha\)) relates to the change in volume (\(V\)) of a substance with a change in temperature (\(\Delta T\)):
\[ \Delta V = V_i \cdot \alpha \cdot \Delta T \]
where \(V_i\) is the initial volume.

### Step 2: Relate volume to density
Density (\(D\)) is defined as mass (\(m\)) divided by volume (\(V\)):
\[ D = \frac{m}{V} \]
Thus, the change in volume will affect the density. If the volume increases, the density decreases.

### Step 3: Express the final and initial densities
Let \(D_i\) be the initial density and \(D_f\) be the final density after the temperature change. The relationship between the initial and final densities can be expressed as:
\[ D_f = \frac{m}{V_f} \]
\[ D_i = \frac{m}{V_i} \]

### Step 4: Substitute volume in terms of density
From the volume expansion formula, we can express the final volume in terms of the initial volume:
\[ V_f = V_i + \Delta V = V_i + V_i \cdot \alpha \cdot \Delta T = V_i(1 + \alpha \Delta T) \]
Now substituting for \(D_f\):
\[ D_f = \frac{m}{V_f} = \frac{m}{V_i(1 + \alpha \Delta T)} = \frac{D_i}{1 + \alpha \Delta T} \]

### Step 5: Find the fractional change in density
The fractional change in density is given by:
\[ \frac{\Delta D}{D_i} = \frac{D_i - D_f}{D_i} = \frac{D_i - \frac{D_i}{1 + \alpha \Delta T}}{D_i} \]
This simplifies to:
\[ \frac{\Delta D}{D_i} = 1 - \frac{1}{1 + \alpha \Delta T} = \frac{\alpha \Delta T}{1 + \alpha \Delta T} \]

### Step 6: Substitute the values
Given:
- \(\alpha = 5 \times 10^{-4} \, K^{-1}\)
- \(\Delta T = 40 \, °C\)

Substituting these values:
\[ \frac{\Delta D}{D_i} = \frac{5 \times 10^{-4} \times 40}{1 + 5 \times 10^{-4} \times 40} \]

### Step 7: Calculate the values
First, calculate the numerator:
\[ 5 \times 10^{-4} \times 40 = 2 \times 10^{-2} \]
Now calculate the denominator:
\[ 1 + 5 \times 10^{-4} \times 40 = 1 + 2 \times 10^{-2} = 1.02 \]
Now, substituting these into the fractional change formula:
\[ \frac{\Delta D}{D_i} = \frac{2 \times 10^{-2}}{1.02} \approx 0.0196078431 \approx 0.0196 \]

### Final Answer
The fractional change in the density of glycerin for a rise of \(40°C\) is approximately:
\[ \frac{\Delta D}{D_i} \approx 1.96 \times 10^{-2} \]