A sample of nitrogen occupies a volume of 350 `cm^3` at STP. Then, its volume at 550 K and 0.5 atm pressure is approximately

Options

A. `1280 cm^3`
B. `1409 cm^3`
C. `1050 cm^3`
D. `1428 cm^3`

Answer

To solve the problem, we will use the combined gas law, which relates the pressure, volume, and temperature of a gas. The formula is given by:

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

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 \(V_1 = 350 \, cm^3\)
- Initial pressure \(P_1 = 1 \, atm\) (at STP)
- Initial temperature \(T_1 = 273 \, K\) (0 degrees Celsius)
- Final pressure \(P_2 = 0.5 \, atm\)
- Final temperature \(T_2 = 550 \, K\)

### Step 2: Rearrange the formula to solve for \(V_2\)
We need to find \(V_2\), so we rearrange the equation:

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

### Step 3: Substitute the known values into the equation
Now we substitute the known values into the rearranged equation:

\[ V_2 = \frac{(1 \, atm) \times (350 \, cm^3) \times (550 \, K)}{(0.5 \, atm) \times (273 \, K)} \]

### Step 4: Calculate \(V_2\)
Now we will perform the calculations step by step:

1. Calculate the numerator:
\[ 1 \times 350 \times 550 = 192500 \, (atm \cdot cm^3 \cdot K) \]

2. Calculate the denominator:
\[ 0.5 \times 273 = 136.5 \, (atm \cdot K) \]

3. Divide the numerator by the denominator:
\[ V_2 = \frac{192500}{136.5} \approx 1409.56 \, cm^3 \]

### Step 5: Round the answer
Rounding to the nearest whole number, we get:
\[ V_2 \approx 1410 \, cm^3 \]

### Final Answer
The volume of nitrogen at 550 K and 0.5 atm pressure is approximately **1410 cm³**.

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