A polythene bag 3 litre capacity is partially filled by 1 liter of Helium gas at 0.4 atm at 300K. Subsequently, enough Ne gas is filled to make total pressure 0.6 atm at 300K. Calculate ratio of moles of Ne to He in the container.

Options

A. `1/2`
B. `2/1`
C. `1/3`
D. `1/4`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of Helium (He)

We can use the Ideal Gas Law, which is given by the formula:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 atm·L/(K·mol))
- \( T \) = temperature in Kelvin

Given:
- Volume of Helium, \( V = 1 \) L (since it is partially filled)
- Pressure of Helium, \( P = 0.4 \) atm
- Temperature, \( T = 300 \) K

Rearranging the formula to find \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the values:

\[ n_{He} = \frac{(0.4 \, \text{atm})(1 \, \text{L})}{(0.0821 \, \text{atm·L/(K·mol)})(300 \, \text{K})} \]

Calculating this gives:

\[ n_{He} = \frac{0.4}{24.63} \approx 0.01624 \, \text{moles} \]

### Step 2: Calculate the total pressure in the bag

The total pressure in the bag after adding Neon gas is given as:

\[ P_{total} = 0.6 \, \text{atm} \]

### Step 3: Calculate the moles of Neon (Ne)

Since the total pressure is the sum of the partial pressures of Helium and Neon, we can express this as:

\[ P_{total} = P_{He} + P_{Ne} \]

Where:
- \( P_{Ne} \) is the pressure contributed by Neon gas.

Using the ideal gas law for the total pressure in the bag:

\[ P_{total} = (n_{He} + n_{Ne}) \frac{RT}{V} \]

We know:
- \( P_{total} = 0.6 \, \text{atm} \)
- \( V = 3 \, \text{L} \)

Substituting the values into the equation:

\[ 0.6 = \left(0.01624 + n_{Ne}\right) \frac{(0.0821)(300)}{3} \]

Calculating \( \frac{(0.0821)(300)}{3} \):

\[ \frac{24.63}{3} = 8.21 \]

Now substituting this back into the equation:

\[ 0.6 = (0.01624 + n_{Ne}) \cdot 8.21 \]

Dividing both sides by 8.21:

\[ \frac{0.6}{8.21} = 0.01624 + n_{Ne} \]

Calculating \( \frac{0.6}{8.21} \):

\[ 0.073 \approx 0.01624 + n_{Ne} \]

Now, solving for \( n_{Ne} \):

\[ n_{Ne} = 0.073 - 0.01624 \approx 0.05676 \, \text{moles} \]

### Step 4: Calculate the ratio of moles of Neon to Helium

Now we can find the ratio of moles of Neon to moles of Helium:

\[ \text{Ratio} = \frac{n_{Ne}}{n_{He}} = \frac{0.05676}{0.01624} \]

Calculating this gives:

\[ \text{Ratio} \approx 3.5 \]

### Final Answer

The ratio of moles of Neon to Helium in the container is approximately:

\[ \text{Ratio} \approx 3.5:1 \]

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