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 moles of gas.

### Step 1: Calculate the initial number of moles of neon gas.
Given:
- Mass of neon gas (m) = 2 g
- Molar mass of neon (M) = 20 g/mol

Using the formula for moles:
\[ n = \frac{m}{M} \]

Substituting the values:
\[ n_1 = \frac{2 \, \text{g}}{20 \, \text{g/mol}} = 0.1 \, \text{mol} \]

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

### Step 3: Determine the new conditions after changes.
New conditions:
- Temperature (T2) = 2 × 300 K = 600 K
- Pressure (P2) = 3 × 3 atm = 9 atm

### Step 4: Use the combined gas law to find the new number of moles (n2).
The combined gas law states:
\[ \frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2} \]

Rearranging gives:
\[ n_2 = \frac{P_2 \cdot n_1 \cdot T_2}{P_1 \cdot T_1} \]

Substituting the known values:
\[ n_2 = \frac{9 \, \text{atm} \cdot 0.1 \, \text{mol} \cdot 600 \, \text{K}}{3 \, \text{atm} \cdot 300 \, \text{K}} \]

Calculating:
\[ n_2 = \frac{9 \cdot 0.1 \cdot 600}{3 \cdot 300} = \frac{540}{900} = 0.6 \, \text{mol} \]

### Step 5: Calculate the mass of neon gas in the new conditions.
Using the formula for mass:
\[ m = n \cdot M \]

Substituting the values:
\[ m_2 = 0.6 \, \text{mol} \cdot 20 \, \text{g/mol} = 12 \, \text{g} \]

### Step 6: Determine the change in mass.
Initial mass (m1) = 2 g
Final mass (m2) = 12 g

Change in mass:
\[ \Delta m = m_2 - m_1 = 12 \, \text{g} - 2 \, \text{g} = 10 \, \text{g} \]

### Conclusion:
To achieve the new conditions, **10 g of neon gas must be added** to the vessel.

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