At N.T.P. the volume of a gas is found to be 273 ml. What will be the volume of this gas at 600 mm Hg and `273^(@)C` ?

Options

A. 391.8 ml
B. 380 ml
C. 691.6 ml
D. 750 ml

Answer

To solve the problem, we will use the combined gas law, which relates pressure, volume, and temperature of a gas. The formula we will use is:

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

### Step 1: Identify the known values
- At NTP (Normal Temperature and Pressure):
- \( P_1 = 1 \, \text{atm} = 760 \, \text{mm Hg} \)
- \( V_1 = 273 \, \text{ml} \)
- \( T_1 = 273 \, \text{°C} = 273 + 273 = 546 \, \text{K} \)

- At the new conditions:
- \( P_2 = 600 \, \text{mm Hg} \)
- \( T_2 = 273 \, \text{°C} = 273 + 273 = 546 \, \text{K} \)
- \( V_2 = ? \)

### Step 2: Convert the temperatures to Kelvin
- The temperature at NTP is given as \( 273 \, \text{°C} \). To convert to Kelvin:

\[ T_1 = 273 + 273 = 546 \, \text{K} \]

- The temperature at the new conditions is also \( 273 \, \text{°C} \), which converts to:

\[ T_2 = 273 + 273 = 546 \, \text{K} \]

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

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

Substituting the values:

\[ \frac{760 \, \text{mm Hg} \times 273 \, \text{ml}}{546 \, \text{K}} = \frac{600 \, \text{mm Hg} \times V_2}{546 \, \text{K}} \]

### Step 4: Simplify the equation
Since \( T_1 \) and \( T_2 \) are the same, we can cancel them out from both sides:

\[ 760 \, \text{mm Hg} \times 273 \, \text{ml} = 600 \, \text{mm Hg} \times V_2 \]

### Step 5: Solve for \( V_2 \)
Now, isolate \( V_2 \):

\[ V_2 = \frac{760 \, \text{mm Hg} \times 273 \, \text{ml}}{600 \, \text{mm Hg}} \]

Calculating this gives:

\[ V_2 = \frac{207480}{600} = 345.8 \, \text{ml} \]

### Step 6: Final answer
Thus, the volume of the gas at 600 mm Hg and 273 °C is approximately:

\[ V_2 \approx 345.8 \, \text{ml} \]