A tank contains 0.85 mol of molecular nitrogen (`N_2`). Determine the mass (in grams) of nitrogen that must be removed from the tank in order to lower the pressure from 38 to 25 atm. Assume that the volume and temperature of the nitrogen in the tank do not change.

Options

A. 2.1 g
B. 8.1g
C. 4.0g
D. 11g

Answer

To solve the problem of determining the mass of nitrogen that must be removed from a tank to lower the pressure from 38 atm to 25 atm, we can use the ideal gas law, which states that \( PV = nRT \). Since the volume (V) and temperature (T) are constant, we can relate the initial and final states of the gas using the pressures and the number of moles.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial pressure, \( P_1 = 38 \, \text{atm} \)
- Final pressure, \( P_2 = 25 \, \text{atm} \)
- Initial number of moles, \( n_1 = 0.85 \, \text{mol} \)

2. **Use the relationship between pressure and number of moles:**
Since \( PV = nRT \) and \( R \) and \( T \) are constants, we can write:
\[ \frac{P_1}{n_1} = \frac{P_2}{n_2} \]
Rearranging gives us:
\[ n_2 = n_1 \cdot \frac{P_2}{P_1} \]

3. **Substitute the known values:**
\[ n_2 = 0.85 \, \text{mol} \cdot \frac{25 \, \text{atm}}{38 \, \text{atm}} \]

4. **Calculate \( n_2 \):**
\[ n_2 = 0.85 \cdot \frac{25}{38} = 0.85 \cdot 0.6579 \approx 0.559 \, \text{mol} \]

5. **Determine the number of moles removed:**
\[ \Delta n = n_1 - n_2 = 0.85 \, \text{mol} - 0.559 \, \text{mol} \approx 0.291 \, \text{mol} \]

6. **Calculate the mass of nitrogen removed:**
The molar mass of nitrogen (\( N_2 \)) is approximately 28 g/mol. Therefore, the mass removed (\( \Delta m \)) can be calculated as:
\[ \Delta m = \Delta n \cdot \text{molar mass} = 0.291 \, \text{mol} \cdot 28 \, \text{g/mol} \]

7. **Perform the calculation:**
\[ \Delta m \approx 0.291 \cdot 28 \approx 8.148 \, \text{g} \]

8. **Final result:**
The mass of nitrogen that must be removed from the tank is approximately **8.15 grams**.