A polythene bag 3 litre capacity is partially filled by 1 liter of Helium gas at 0.2 atm at 300K. Subsequently, enough Ne gas is filled to make total pressure 0.4 atm at 300K. Calculate ratio of moles of Ne to He in the container.

Options

A. `1`
B. `1/4`
C. `1/3`
D. `1/5`

Explanation

To solve the problem, we need to calculate the number of moles of Helium (He) and Neon (Ne) gases in the polythene bag and then find the ratio of moles of Ne to He.

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

We use the ideal gas law equation:

\[ PV = nRT \]

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 the bag = 3 L
- Volume of He = 1 L (since it is partially filled)
- Pressure of He = 0.2 atm
- Temperature = 300 K

Using the ideal gas law for Helium:

\[ 0.2 \, \text{atm} \times 1 \, \text{L} = n_{\text{He}} \times 0.0821 \, \text{atm·L/(K·mol)} \times 300 \, \text{K} \]

Now, rearranging for \( n_{\text{He}} \):

\[ n_{\text{He}} = \frac{0.2 \times 1}{0.0821 \times 300} \]

Calculating:

\[ n_{\text{He}} = \frac{0.2}{24.63} \approx 0.00812 \, \text{moles} \]

### Step 2: Calculate the total moles of gas when Ne is added

After adding Neon gas, the total pressure in the bag becomes 0.4 atm. We can use the total pressure to find the moles of Neon.

Let \( n_{\text{Ne}} \) be the moles of Neon gas. The total moles of gas in the container will be:

\[ n_{\text{total}} = n_{\text{He}} + n_{\text{Ne}} = 0.00812 + n_{\text{Ne}} \]

Using the ideal gas law for the total pressure:

\[ 0.4 \, \text{atm} \times 3 \, \text{L} = (0.00812 + n_{\text{Ne}}) \times 0.0821 \times 300 \]

Rearranging gives:

\[ 0.4 \times 3 = (0.00812 + n_{\text{Ne}}) \times 24.63 \]

Calculating the left side:

\[ 1.2 = (0.00812 + n_{\text{Ne}}) \times 24.63 \]

Now, divide both sides by 24.63:

\[ \frac{1.2}{24.63} = 0.00812 + n_{\text{Ne}} \]

Calculating:

\[ 0.0488 \approx 0.00812 + n_{\text{Ne}} \]

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

\[ n_{\text{Ne}} \approx 0.0488 - 0.00812 \approx 0.04068 \, \text{moles} \]

### Step 3: Calculate the ratio of moles of Ne to He

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

\[ \text{Ratio} = \frac{n_{\text{Ne}}}{n_{\text{He}}} = \frac{0.04068}{0.00812} \approx 5 \]

### Final Answer

The ratio of moles of Neon to Helium in the container is approximately **5:1**.

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