A closed container of volume `0.02m^3`contains a mixture of neon and argon gases, at a temperature of `27^@C` and pressure of `1xx10^5Nm^-2`. The total mass of the mixture is 28g. If the molar masses of neon and argon are `20 and 40gmol^-1` respectively, find the masses of the individual gasses in the container assuming them to be ideal (Universal gas constant `R=8.314J//mol-K`).

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
We have the following information:
- Volume of the container, \( V = 0.02 \, m^3 \)
- Temperature, \( T = 27^\circ C = 27 + 273 = 300 \, K \)
- Pressure, \( P = 1 \times 10^5 \, N/m^2 \)
- Total mass of the gas mixture, \( m_{total} = 28 \, g = 0.028 \, kg \)
- Molar mass of Neon, \( M_{Ne} = 20 \, g/mol = 0.020 \, kg/mol \)
- Molar mass of Argon, \( M_{Ar} = 40 \, g/mol = 0.040 \, kg/mol \)

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( n \) is the total number of moles of the gas mixture.
- \( R = 8.314 \, J/(mol \cdot K) \)

Rearranging the equation to find \( n \):
\[ n = \frac{PV}{RT} \]

### Step 3: Calculate the Total Number of Moles
Substituting the known values into the equation:
\[ n = \frac{(1 \times 10^5 \, N/m^2)(0.02 \, m^3)}{(8.314 \, J/(mol \cdot K))(300 \, K)} \]

Calculating the numerator:
\[ 1 \times 10^5 \times 0.02 = 2000 \, N \cdot m = 2000 \, J \]

Calculating the denominator:
\[ 8.314 \times 300 = 2494.2 \, J/(mol \cdot K) \]

Now substituting these values:
\[ n = \frac{2000}{2494.2} \approx 0.802 \, mol \]

### Step 4: Set Up the Mass Equations
Let \( m_{Ne} \) be the mass of Neon and \( m_{Ar} \) be the mass of Argon. We know:
\[ m_{Ne} + m_{Ar} = 28 \, g \]

Also, we can express the number of moles in terms of mass:
\[ n = \frac{m_{Ne}}{M_{Ne}} + \frac{m_{Ar}}{M_{Ar}} \]

Substituting the molar masses:
\[ n = \frac{m_{Ne}}{0.020} + \frac{m_{Ar}}{0.040} \]

### Step 5: Substitute and Solve
From the first equation, we can express \( m_{Ar} \):
\[ m_{Ar} = 28 - m_{Ne} \]

Substituting this into the moles equation:
\[ 0.802 = \frac{m_{Ne}}{0.020} + \frac{28 - m_{Ne}}{0.040} \]

Multiplying through by 40 to eliminate the denominators:
\[ 0.802 \times 40 = 2m_{Ne} + (28 - m_{Ne}) \]
\[ 32.08 = 2m_{Ne} + 28 - m_{Ne} \]
\[ 32.08 - 28 = m_{Ne} \]
\[ m_{Ne} = 4.08 \, g \]

### Step 6: Calculate Mass of Argon
Using \( m_{Ne} \) to find \( m_{Ar} \):
\[ m_{Ar} = 28 - 4.08 = 23.92 \, g \]

### Final Answer
- Mass of Neon, \( m_{Ne} \approx 4.08 \, g \)
- Mass of Argon, \( m_{Ar} \approx 23.92 \, g \)