A 0.50 L container is ocupied by nitrogen at a pressure of 800 torr and a temperature of `0^(@)C`. The container can only withstand a pressure of 3.0 atm. What is the highest temperature `("^(@)C)` to which the container may be heated?

Options

A. 505
B. 450
C. 625
D. 560

Explanation

To solve the problem, we will use the relationship between pressure and temperature for a gas at constant volume, which can be derived from the ideal gas law. Here's a step-by-step solution:

### Step 1: Identify the given values
- Volume (V) = 0.50 L (not needed for calculations since it's constant)
- Initial Pressure (P1) = 800 torr
- Maximum Pressure (P2) = 3.0 atm
- Initial Temperature (T1) = 0°C = 273 K (convert to Kelvin)

### Step 2: Convert P1 from torr to atm
To convert the pressure from torr to atm, we use the conversion factor:
1 atm = 760 torr

\[ P1 = \frac{800 \text{ torr}}{760 \text{ torr/atm}} \approx 1.05 \text{ atm} \]

### Step 3: Use the relationship between pressure and temperature
Since the volume and the number of moles of gas are constant, we can use the formula that relates pressure and temperature:

\[ \frac{P1}{P2} = \frac{T1}{T2} \]

Where:
- \(P1\) = initial pressure
- \(P2\) = maximum pressure
- \(T1\) = initial temperature in Kelvin
- \(T2\) = final temperature in Kelvin (which we want to find)

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:

\[ \frac{1.05 \text{ atm}}{3.0 \text{ atm}} = \frac{273 \text{ K}}{T2} \]

### Step 5: Solve for T2
Cross-multiplying to solve for \(T2\):

\[ 1.05 \times T2 = 3.0 \times 273 \]

Calculating the right side:

\[ 1.05 \times T2 = 819 \]

Now, divide both sides by 1.05:

\[ T2 = \frac{819}{1.05} \approx 780 \text{ K} \]

### Step 6: Convert T2 from Kelvin to Celsius
To convert from Kelvin to Celsius:

\[ T2 = 780 \text{ K} - 273 = 507 \text{ °C} \]

### Final Answer
The highest temperature to which the container may be heated is approximately **507 °C**.

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