A vessel of volume `2 xx 10^(-2) m^(3)` contains a mixture of hydrogen and helium at `47^(@)C` temperature and `4.15 xx 10^(5) N//m^(2)` Pressure. The mass of the mixture is `10^(-2) kg`. Calculate the masses of hydrogen and helium in the given mixture.

Explanation

To solve the problem of finding the masses of hydrogen and helium in the given mixture, we will use the ideal gas equation and some algebraic manipulations. Here’s a step-by-step breakdown of the solution:

### Step 1: Write down the known values
- Volume \( V = 2 \times 10^{-2} \, m^3 \)
- Temperature \( T = 47^\circ C = 47 + 273 = 320 \, K \)
- Pressure \( P = 4.15 \times 10^5 \, N/m^2 \)
- Total mass of the mixture \( m_1 + m_2 = 10^{-2} \, kg = 10 \, g \)

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