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 0.5 atm at 300K. Calculate ratio of moles of Ne to He in the container.

Options

A. `1/6`
B. `5/6`
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 equation, which is given by:

\[ PV = nRT \]

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

Given:
- Volume of the bag, \( V = 3 \) liters
- Pressure of Helium, \( P_{He} = 0.4 \) atm
- Temperature, \( T = 300 \) K

Rearranging the ideal gas law to find \( n \):

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

Substituting the values for Helium:

\[ n_{He} = \frac{0.4 \, \text{atm} \times 3 \, \text{L}}{0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}} \]

Calculating:

\[ n_{He} = \frac{1.2}{24.63} \approx 0.0487 \, \text{moles} \]

### Step 2: Calculate the total pressure and moles of Neon (Ne)

The total pressure in the bag after adding Neon is given as \( P_{total} = 0.5 \) atm.

Using the ideal gas law again for the total pressure:

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

Substituting the known values:

\[ 0.5 = (0.0487 + n_{Ne}) \frac{0.0821 \times 300}{3} \]

Calculating \( \frac{RT}{V} \):

\[ \frac{0.0821 \times 300}{3} = 8.21 \]

Now substituting this back into the equation:

\[ 0.5 = (0.0487 + n_{Ne}) \times 8.21 \]

Now we can solve for \( n_{Ne} \):

\[ 0.5 = 0.0487 \times 8.21 + n_{Ne} \times 8.21 \]

Calculating \( 0.0487 \times 8.21 \):

\[ 0.5 = 0.3997 + n_{Ne} \times 8.21 \]

Subtracting \( 0.3997 \) from both sides:

\[ 0.5 - 0.3997 = n_{Ne} \times 8.21 \]

\[ 0.1003 = n_{Ne} \times 8.21 \]

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

\[ n_{Ne} = \frac{0.1003}{8.21} \approx 0.0122 \, \text{moles} \]

### Step 3: Calculate the ratio of moles of Neon to Helium

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

\[ \text{Ratio} = \frac{n_{Ne}}{n_{He}} = \frac{0.0122}{0.0487} \]

Calculating the ratio:

\[ \text{Ratio} \approx 0.2505 \]

This can be approximated as:

\[ \text{Ratio} \approx \frac{1}{4} \]

### Final Answer

The ratio of moles of Neon to Helium in the container is approximately \( 1:4 \).

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