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 `253^o`C?

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:
- \( P \) = Pressure
- \( V \) = Volume
- \( T \) = Temperature (in Kelvin)

### Step 1: Identify the given values
- Volume at NTP (\( V_1 \)) = 270 mL
- Pressure at NTP (\( P_1 \)) = 1 atm
- Temperature at NTP (\( T_1 \)) = 293 K (which is \( 0^\circ C + 273 \))
- New pressure (\( P_2 \)) = 600 mm Hg
- New temperature (\( T_2 \)) = 253 °C

### Step 2: Convert the new pressure to atm
To convert mm Hg to atm, we use the conversion factor:
\[ 1 \text{ atm} = 760 \text{ mm Hg} \]
Thus,
\[ P_2 = \frac{600 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.7895 \text{ atm} \approx 0.79 \text{ atm} \]

### Step 3: Convert the new temperature to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T_2 = 253 + 273 = 526 \text{ K} \]

### Step 4: Substitute the values into the Ideal Gas Law equation
Now we can substitute the known values into the Ideal Gas Law equation:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Substituting the values:
\[ \frac{1 \text{ atm} \times 270 \text{ mL}}{293 \text{ K}} = \frac{0.79 \text{ atm} \times V_2}{526 \text{ K}} \]

### Step 5: Solve for \( V_2 \)
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{1 \text{ atm} \times 270 \text{ mL} \times 526 \text{ K}}{0.79 \text{ atm} \times 293 \text{ K}} \]

Calculating the right side:
\[ V_2 = \frac{270 \times 526}{0.79 \times 293} \]

Calculating the numerator:
\[ 270 \times 526 = 142020 \]

Calculating the denominator:
\[ 0.79 \times 293 = 231.07 \]

Now, dividing the two results:
\[ V_2 = \frac{142020}{231.07} \approx 613.55 \text{ mL} \]

### Final Answer
The volume of the gas at 600 mm Hg and 253 °C is approximately **613.55 mL**.

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