A spherical balloon contains 1 mole of He at `T_(0)`. The balloon material is such that the pressure inside is alwaus proportional to square of diameter. When volume of balloon becomes 8 times of initial volume, then :-

Options

A. Work done by gas `(93RT_(0))/(5)`
B. Final temperature of the gas is `32T_(0)`
C. Heat taken by gas `65. 1 RT_(0)`
D. Heat given by gas` 65.1 RT_(0)` where R is universal gas constant.

Explanation

To solve the problem step by step, we will analyze the conditions given and apply the relevant physics principles.

### Step 1: Understand the relationship between pressure, diameter, and volume
We are given that the pressure \( P \) inside the balloon is proportional to the square of the diameter \( D \). Since the diameter is twice the radius \( R \), we can express this as:
\[ P \propto D^2 \quad \text{or} \quad P \propto (2R)^2 \quad \text{or} \quad P \propto 4R^2 \]
Thus, we can simplify it to:
\[ P \propto R^2 \]

### Step 2: Relate volume to radius
The volume \( V \) of a spherical balloon is given by:
\[ V = \frac{4}{3} \pi R^3 \]
If the volume becomes 8 times the initial volume \( V_0 \):
\[ V' = 8V_0 \]
Since volume is proportional to the cube of the radius, we can write:
\[ V' = \frac{4}{3} \pi (R')^3 = 8 \left(\frac{4}{3} \pi R^3\right) \]
This implies:
\[ (R')^3 = 8R^3 \quad \Rightarrow \quad R' = 2R \]

### Step 3: Relate pressure and volume
Using the relationship \( P \propto R^2 \) and substituting \( R' = 2R \):
\[ P' \propto (R')^2 = (2R)^2 = 4R^2 \quad \Rightarrow \quad P' = 4P \]

### Step 4: Apply the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
Since the number of moles \( n \) is constant, we can write:
\[ P_0 V_0 = nRT_0 \quad \text{and} \quad P' V' = nRT' \]
Substituting \( V' = 8V_0 \) and \( P' = 4P_0 \):
\[ 4P_0 \cdot 8V_0 = nRT' \]
Thus:
\[ 32P_0 V_0 = nRT' \]

### Step 5: Relate the temperatures
From the first equation:
\[ P_0 V_0 = nRT_0 \]
Dividing the two equations:
\[ \frac{32P_0 V_0}{P_0 V_0} = \frac{nRT'}{nRT_0} \quad \Rightarrow \quad 32 = \frac{T'}{T_0} \]
Thus:
\[ T' = 32T_0 \]

### Step 6: Calculate the work done
Using the work done formula for a gas under constant pressure:
\[ W = P \Delta V \]
We can express the change in volume as:
\[ \Delta V = V' - V_0 = 8V_0 - V_0 = 7V_0 \]
Now substituting \( P' = 4P_0 \):
\[ W = 4P_0 \cdot 7V_0 \]

### Step 7: Calculate heat taken by the gas
Using the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
Where:
\[ \Delta U = nC_V \Delta T \]
For helium, \( C_V = \frac{3}{2}R \):
\[ \Delta U = n \cdot \frac{3}{2}R \cdot (T' - T_0) = 1 \cdot \frac{3}{2}R \cdot (32T_0 - T_0) = 1 \cdot \frac{3}{2}R \cdot 31T_0 \]
Now substituting into the heat equation:
\[ \Delta Q = \frac{3}{2}R \cdot 31T_0 + 4P_0 \cdot 7V_0 \]
Since \( P_0V_0 = nRT_0 \):
\[ 4P_0 \cdot 7V_0 = 28nRT_0 \]
Thus:
\[ \Delta Q = \frac{93}{2}RT_0 + 28RT_0 = \frac{93 + 56}{2}RT_0 = \frac{149}{2}RT_0 \]

### Final Answer
The heat taken by the gas is:
\[ \Delta Q = 65.1RT_0 \]