A gas occupies 300 ml at `27^(@)C` and 740 mm Hg pressure. Calculate its volume at S.T.P. ? 

Options

A. 0.3650 L
B. 0.2658 L
C. 200 L
D. 365 L

Explanation

To solve the problem of finding the volume of a gas at Standard Temperature and Pressure (S.T.P) given its initial conditions, we can use the combined gas law, which states that:

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

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature in Kelvin
- \( P_2 \) = final pressure (at S.T.P)
- \( V_2 \) = final volume (at S.T.P)
- \( T_2 \) = final temperature (at S.T.P)

### Step 1: Convert the initial temperature to Kelvin
The initial temperature is given as \( 27^\circ C \). To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T_1 = 27 + 273.15 = 300.15 \, K \]

### Step 2: Identify the values
From the problem, we have:
- \( V_1 = 300 \, ml \)
- \( P_1 = 740 \, mm \, Hg \)
- \( T_1 = 300.15 \, K \)
- At S.T.P, \( P_2 = 760 \, mm \, Hg \) and \( T_2 = 273.15 \, K \)

### Step 3: Substitute the values into the combined gas law
Now we can substitute the known values into the combined gas law:

\[ \frac{740 \, mm \, Hg \times 300 \, ml}{300.15 \, K} = \frac{760 \, mm \, Hg \times V_2}{273.15 \, K} \]

### Step 4: Rearrange the equation to solve for \( V_2 \)
Rearranging the equation to isolate \( V_2 \):

\[ V_2 = \frac{740 \, mm \, Hg \times 300 \, ml \times 273.15 \, K}{760 \, mm \, Hg \times 300.15 \, K} \]

### Step 5: Calculate \( V_2 \)
Now we can calculate \( V_2 \):

\[ V_2 = \frac{740 \times 300 \times 273.15}{760 \times 300.15} \]

Calculating the numerator:

\[ 740 \times 300 \times 273.15 = 607,785,300 \]

Calculating the denominator:

\[ 760 \times 300.15 = 228,114 \]

Now, dividing the two results:

\[ V_2 = \frac{607785300}{228114} \approx 266.8 \, ml \]

### Final Answer
The volume of the gas at S.T.P is approximately \( 266.8 \, ml \).

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