At N.T.P the volume of a gas is found to be 270mL. What will be the volume of this gas at 600mm Hg and `173^o`C?

Options

A. 245
B. 520
C. 255
D. 458

Explanation

To find the volume of the gas at the given conditions, we can use the combined gas law, which relates pressure, volume, and temperature. The formula is:

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

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature
- \( P_2 \) = final pressure
- \( V_2 \) = final volume
- \( T_2 \) = final temperature

### Step 1: Identify the known values
- Initial volume \( V_1 = 270 \, \text{mL} \)
- Initial pressure \( P_1 = 1 \, \text{atm} \) (NTP)
- Initial temperature \( T_1 = 293.15 \, \text{K} \) (NTP)
- Final pressure \( P_2 = 600 \, \text{mm Hg} \)
- Final temperature \( T_2 = 173 \, \text{°C} \)

### Step 2: Convert the final pressure to atm
\[ P_2 = \frac{600 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.7895 \, \text{atm} \]

### Step 3: Convert the final temperature to Kelvin
\[ T_2 = 173 \, \text{°C} + 273.15 = 446.15 \, \text{K} \]

### Step 4: Substitute the known values into the combined gas law
Using the formula:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Substituting the values:
\[ \frac{(1 \, \text{atm}) (270 \, \text{mL})}{293.15 \, \text{K}} = \frac{(0.7895 \, \text{atm}) V_2}{446.15 \, \text{K}} \]

### Step 5: Rearranging to solve for \( V_2 \)
\[ V_2 = \frac{(1 \, \text{atm}) (270 \, \text{mL}) (446.15 \, \text{K})}{(0.7895 \, \text{atm}) (293.15 \, \text{K})} \]

### Step 6: Calculate \( V_2 \)
Calculating the right side:
\[ V_2 = \frac{270 \times 446.15}{0.7895 \times 293.15} \]
\[ V_2 = \frac{120,000.5}{231.12} \approx 519.5 \, \text{mL} \]

### Step 7: Round to appropriate significant figures
The final volume \( V_2 \approx 520 \, \text{mL} \).

### Final Answer
The volume of the gas at 600 mm Hg and 173 °C is approximately **520 mL**.

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