For a closed container containing `100` mol of an ideal gas fitted with movable, frictionless, weightless piston operating such that pressure of gas remain constant at `8.21` atm, which graph repsents correct variation of log `V v//s` log `T` where `V` is in litre and `T` is in Kelvin ? `(R = 0.0821 (atmL)/(molK))`

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the relationship between the volume (V) and temperature (T) of an ideal gas under constant pressure conditions. We will use the ideal gas law and logarithmic properties to derive the relationship.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 100 mol
- Pressure (P) = 8.21 atm
- Gas constant (R) = 0.0821 (atm·L)/(mol·K)

2. **Use the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
Rearranging this equation for volume (V):
\[ V = \frac{nRT}{P} \]

3. **Substituting the Known Values:**
Substitute the known values into the equation:
\[ V = \frac{100 \, \text{mol} \times 0.0821 \, \frac{\text{atm·L}}{\text{mol·K}} \times T}{8.21 \, \text{atm}} \]
Simplifying this gives:
\[ V = \frac{100 \times 0.0821 \times T}{8.21} \]

4. **Calculate the Constant:**
Calculate the constant factor:
\[ V = \frac{8.21}{8.21} \times 100 \times 0.0821 \times T = 100 \times 0.010 \times T = 0.821 \times T \]

5. **Taking Logarithm:**
Now, take the logarithm of both sides:
\[ \log V = \log(0.821 \times T) \]
Using the logarithmic property \(\log(ab) = \log a + \log b\):
\[ \log V = \log 0.821 + \log T \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ \log V = \log T + \log 0.821 \]
This indicates a linear relationship between \(\log V\) and \(\log T\).

7. **Graph Interpretation:**
The graph of \(\log V\) vs. \(\log T\) will be a straight line with a slope of 1, indicating that as temperature increases, volume also increases at a constant pressure.

### Conclusion:
The correct graph representing the variation of \(\log V\) vs. \(\log T\) is a straight line with a positive slope.