At N.T.P the volume of a gas is found to be 250mL. What will be the volume of this gas at 600mm Hg and `250^o`C?

Options

A. 358.79
B. 478.58
C. 255.78
D. 564.86

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that:

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

Where:
- \( P_1 \) and \( V_1 \) are the pressure and volume of the gas at the initial conditions (NTP),
- \( T_1 \) is the initial temperature,
- \( P_2 \) and \( V_2 \) are the pressure and volume of the gas at the final conditions,
- \( T_2 \) is the final temperature.

### Step 1: Identify the initial conditions (NTP)
At Normal Temperature and Pressure (NTP):
- \( V_1 = 250 \, \text{mL} \)
- \( P_1 = 1 \, \text{atm} \)
- \( T_1 = 293 \, \text{K} \) (which is 0°C)

### Step 2: Identify the final conditions
Given:
- \( P_2 = 600 \, \text{mm Hg} \)
- Convert \( P_2 \) to atm:
\[ P_2 = \frac{600 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.789 \, \text{atm} \]

- \( T_2 = 250 \, \text{°C} \)
- Convert \( T_2 \) to Kelvin:
\[ T_2 = 250 + 273 = 523 \, \text{K} \]

### Step 3: Substitute values into the Ideal Gas Law
Using the equation:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Substituting the known values:
\[ \frac{1 \, \text{atm} \times 250 \, \text{mL}}{293 \, \text{K}} = \frac{0.789 \, \text{atm} \times V_2}{523 \, \text{K}} \]

### Step 4: Solve for \( V_2 \)
Cross-multiplying gives:
\[ 1 \times 250 \times 523 = 0.789 \times V_2 \times 293 \]
\[ 130750 = 231.357 \times V_2 \]
Now, solve for \( V_2 \):
\[ V_2 = \frac{130750}{231.357} \approx 564.86 \, \text{mL} \]

### Final Answer
The volume of the gas at 600 mm Hg and 250°C is approximately **564.86 mL**.

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