If 2g neon gas is present in a vessel at 3 atm and 300 K and temperature is doubled and pressure becomes triple then the mass of neon that must be added or subtracted must be

Options

A. 1 g neon must be removed
B. 1 g neon must be added
C. 2g neon must be added
D. 3g neon must be removed

Answer

To solve the problem step by step, we will use the ideal gas law and the relationship between pressure, volume, temperature, and the number of moles of gas.

### Step 1: Determine Initial Moles of Neon
Given:
- Mass of Neon (m) = 2 g
- Molar mass of Neon (M) = 20 g/mol

To find the number of moles (n):
\[ n = \frac{m}{M} = \frac{2 \, \text{g}}{20 \, \text{g/mol}} = 0.1 \, \text{mol} \]

**Hint for Step 1:** Use the formula for moles, which is the mass of the gas divided by its molar mass.

### Step 2: Identify Initial Conditions
Initial conditions:
- Pressure (P1) = 3 atm
- Temperature (T1) = 300 K

### Step 3: Determine New Conditions
New conditions:
- Temperature is doubled: \( T2 = 2 \times 300 \, \text{K} = 600 \, \text{K} \)
- Pressure is tripled: \( P2 = 3 \times 3 \, \text{atm} = 9 \, \text{atm} \)

### Step 4: Use the Combined Gas Law
The combined gas law states:
\[ \frac{P_1 n_1}{T_1} = \frac{P_2 n_2}{T_2} \]
Substituting known values:
\[ \frac{3 \, \text{atm} \times 0.1 \, \text{mol}}{300 \, \text{K}} = \frac{9 \, \text{atm} \times n_2}{600 \, \text{K}} \]

### Step 5: Solve for \( n_2 \)
Cross-multiplying gives:
\[ 3 \times 0.1 \times 600 = 9 \times n_2 \times 300 \]
\[ 180 = 2700 n_2 \]
\[ n_2 = \frac{180}{2700} = 0.06667 \, \text{mol} \]

### Step 6: Calculate the Mass of Neon at New Conditions
Now, calculate the mass of Neon at the new conditions:
\[ \text{Mass} = n_2 \times M = 0.06667 \, \text{mol} \times 20 \, \text{g/mol} = 1.3334 \, \text{g} \]

### Step 7: Determine the Change in Mass
Initial mass = 2 g, Final mass = 1.3334 g
Change in mass:
\[ \text{Change} = \text{Initial mass} - \text{Final mass} = 2 \, \text{g} - 1.3334 \, \text{g} = 0.6666 \, \text{g} \]

### Step 8: Conclusion
Since the final mass is less than the initial mass, we need to subtract approximately 0.67 g of Neon.

**Final Answer:** Approximately 0.67 g of Neon must be subtracted.

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