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 of finding the volume of nitrogen gas at a new temperature and pressure, we will use the combined gas law. The combined gas law 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 need to find)
- \( T_2 \) = final temperature

### Step 1: Identify the given values
- Initial pressure, \( P_1 = 1 \) atm (at STP)
- Initial volume, \( V_1 = 350 \) cm³
- Initial temperature, \( T_1 = 273 \) K (standard temperature)
- Final pressure, \( P_2 = 0.5 \) atm
- Final temperature, \( T_2 = 550 \) K

### Step 2: Substitute the known values into the combined gas law equation
We can rearrange the combined gas law to solve for \( V_2 \):

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

Substituting the known values:

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

### Step 3: Calculate the volume \( V_2 \)
Now, we can calculate \( V_2 \):

1. Calculate the numerator:
\[ 1 \times 350 \times 550 = 192500 \]

2. Calculate the denominator:
\[ 0.5 \times 273 = 136.5 \]

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

### Step 4: Round the answer
The volume \( V_2 \) can be approximated to 1409 cm³.

### Final Answer
Thus, the volume of nitrogen at 550 K and 0.5 atm pressure is approximately **1409 cm³**.

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