A gas is heated in such a way that its pressure and volume both become double. Now by decreasing temperature, some of air molecules were introduced into the container to maintain the increased volume and pressure. Assuming `1//4^(th)` of the initial number of moles has been given for this purpose. By what fraction of temperature has been raised finally of initial absolute temperature.

Options

A. 4 times
B. `(16)/(5)` times
C. `(4)/(5)`times
D. `(1)/(5)`times

Answer

To solve the problem step by step, we will use the ideal gas law and the information provided in the question.

### Step 1: Understand the initial conditions
Let the initial pressure, volume, and temperature of the gas be \( P_0 \), \( V_0 \), and \( T_0 \) respectively. According to the ideal gas law:
\[ P_0 V_0 = n R T_0 \]

### Step 2: Determine the new conditions after heating
After heating, both the pressure and volume are doubled:
\[ P' = 2P_0 \quad \text{and} \quad V' = 2V_0 \]
Let the new temperature after heating be \( T' \). Using the ideal gas law again:
\[ P' V' = n R T' \]
Substituting the new values:
\[ (2P_0)(2V_0) = n R T' \]
This simplifies to:
\[ 4P_0 V_0 = n R T' \]

### Step 3: Relate the new temperature to the initial temperature
From the initial condition, we have:
\[ P_0 V_0 = n R T_0 \]
Substituting this into the equation for \( T' \):
\[ 4(n R T_0) = n R T' \]
Dividing both sides by \( nR \):
\[ T' = 4T_0 \]

### Step 4: Introduce additional moles of gas
Now, we introduce \( \frac{1}{4} \) of the initial number of moles into the container. The new number of moles becomes:
\[ n' = n + \frac{n}{4} = \frac{5n}{4} \]

### Step 5: Determine the new temperature after adding moles
Let the new temperature after adding the moles be \( T'' \). The pressure and volume are maintained at \( 2P_0 \) and \( 2V_0 \):
\[ P' V' = n' R T'' \]
Substituting the values:
\[ (2P_0)(2V_0) = \left(\frac{5n}{4}\right) R T'' \]
This simplifies to:
\[ 4P_0 V_0 = \left(\frac{5n}{4}\right) R T'' \]
Using \( P_0 V_0 = n R T_0 \):
\[ 4(n R T_0) = \left(\frac{5n}{4}\right) R T'' \]
Dividing both sides by \( nR \):
\[ 4T_0 = \frac{5}{4} T'' \]

### Step 6: Solve for \( T'' \)
Multiplying both sides by 4:
\[ 16T_0 = 5T'' \]
Now, solving for \( T'' \):
\[ T'' = \frac{16T_0}{5} \]

### Step 7: Find the fraction of the temperature raised
The initial temperature is \( T_0 \) and the final temperature is \( T'' \):
\[ \text{Fraction of temperature raised} = \frac{T'' - T_0}{T_0} = \frac{\frac{16T_0}{5} - T_0}{T_0} \]
This simplifies to:
\[ = \frac{\frac{16T_0}{5} - \frac{5T_0}{5}}{T_0} = \frac{\frac{11T_0}{5}}{T_0} = \frac{11}{5} \]

### Final Answer
The fraction of the temperature raised finally from the initial absolute temperature is:
\[ \frac{11}{5} \]