At `27^(@)C` and under one atmosphere pressure, a gas occupies a volume of V L. In case the temperature is increased to `177^(@)C` and pressure to `1.5` bar, the corresponding volume will be :

Options

A. `V mL`
B. `2V mL`
C. `V//2 mL`
D. `V//3 mL`

Explanation

To solve the problem, we will use the ideal gas law in the form of the combined gas law, which states:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step 1: Identify the given values
- Initial pressure, \( P_1 = 1 \, \text{atm} \) (which we can approximate as \( 1 \, \text{bar} \))
- Initial volume, \( V_1 = V \, \text{L} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
- Final pressure, \( P_2 = 1.5 \, \text{bar} \)
- Final temperature, \( T_2 = 177^\circ C = 177 + 273 = 450 \, \text{K} \)

### Step 2: Rearrange the combined gas law to solve for \( V_2 \)
We need to isolate \( V_2 \) in the equation:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Rearranging gives:

\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 3: Substitute the known values into the equation
Now, we substitute the known values into the equation:

\[ V_2 = \frac{(1 \, \text{bar}) (V \, \text{L}) (450 \, \text{K})}{(1.5 \, \text{bar}) (300 \, \text{K})} \]

### Step 4: Simplify the equation
Now we simplify the equation step by step:

\[ V_2 = \frac{1 \cdot V \cdot 450}{1.5 \cdot 300} \]

Calculating the denominator:

\[ 1.5 \cdot 300 = 450 \]

So now we have:

\[ V_2 = \frac{450 V}{450} \]

### Step 5: Final calculation
The \( 450 \) cancels out:

\[ V_2 = V \, \text{L} \]

### Conclusion
Thus, the corresponding volume \( V_2 \) at the new conditions is:

\[ \boxed{V \, \text{L}} \]