At N.T.P the volume of a gas is found to be 237mL. What will be the volume of this gas at 600mm Hg and `237^o`C?

Explanation

To solve the problem of finding the volume of a gas at different conditions, we can use the Ideal Gas Law, which states that:

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

Where:
- \(P_1\) = Initial pressure
- \(V_1\) = Initial volume
- \(T_1\) = Initial temperature
- \(P_2\) = Final pressure
- \(V_2\) = Final volume (what we want to find)
- \(T_2\) = Final temperature

### Step 1: Identify the known values
- At NTP (Normal Temperature and Pressure):
- \(V_1 = 237 \, \text{mL}\)
- \(P_1 = 1 \, \text{atm}\)
- \(T_1 = 293.15 \, \text{K}\) (which is 0°C)

- Final conditions:
- \(P_2 = 600 \, \text{mm Hg}\)
- \(T_2 = 227 \, \text{°C} = 227 + 273.15 = 500.15 \, \text{K}\)

### Step 2: Convert the pressure from mm Hg to atm
To convert pressure from mm Hg to atm, we use the conversion factor:
\[ 1 \, \text{atm} = 760 \, \text{mm Hg} \]
Thus,
\[ P_2 = \frac{600 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.789 \, \text{atm} \]

### Step 3: Substitute the known values into the Ideal Gas Law equation
Now we can substitute the known values into the equation:
\[ \frac{1 \, \text{atm} \cdot 237 \, \text{mL}}{293.15 \, \text{K}} = \frac{0.789 \, \text{atm} \cdot V_2}{500.15 \, \text{K}} \]

### Step 4: Rearranging to find \(V_2\)
Rearranging the equation to solve for \(V_2\):
\[ V_2 = \frac{1 \, \text{atm} \cdot 237 \, \text{mL} \cdot 500.15 \, \text{K}}{0.789 \, \text{atm} \cdot 293.15 \, \text{K}} \]

### Step 5: Calculate \(V_2\)
Now we can calculate \(V_2\):
\[ V_2 = \frac{1 \cdot 237 \cdot 500.15}{0.789 \cdot 293.15} \approx \frac{118500.55}{231.1} \approx 513.5 \, \text{mL} \]

### Final Answer
The volume of the gas at 600 mm Hg and 227 °C is approximately **513.5 mL**.

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