Two containers of equal volume enclose equal masses of hydrogen and helium at `300K`. Find (i) the ratio of pressure and (ii) the ratio of total `K.E.` of all the molecules of the two gases.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- We have two containers of equal volume containing equal masses of hydrogen (H₂) and helium (He) at a temperature of 300 K.
- The molecular mass of hydrogen (H₂) is 2 g/mol, and the molecular mass of helium (He) is 4 g/mol.

### Step 2: Find the ratio of pressure
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

Since the volume, temperature, and gas constant are the same for both gases, we can express the number of moles \( n \) as:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass of the gas
- \( M \) = molar mass of the gas

Thus, the pressure can be expressed as:
\[ P = \frac{m}{MV}RT \]

For hydrogen:
\[ P_{H_2} = \frac{m}{2V}RT \]

For helium:
\[ P_{He} = \frac{m}{4V}RT \]

Now, we can find the ratio of pressures:
\[ \frac{P_{H_2}}{P_{He}} = \frac{\frac{m}{2V}RT}{\frac{m}{4V}RT} = \frac{4}{2} = 2 \]

### Step 3: Ratio of total kinetic energy
The total kinetic energy (KE) of a gas can be expressed as:
\[ KE = \frac{F}{2} nRT \]
Where \( F \) is the degrees of freedom.

For hydrogen (a diatomic gas):
- Degrees of freedom \( F = 5 \)
- Therefore, the kinetic energy for hydrogen is:
\[ KE_{H_2} = \frac{5}{2} n_{H_2} RT \]

For helium (a monatomic gas):
- Degrees of freedom \( F = 3 \)
- Therefore, the kinetic energy for helium is:
\[ KE_{He} = \frac{3}{2} n_{He} RT \]

Now substituting \( n \) in terms of mass:
For hydrogen:
\[ KE_{H_2} = \frac{5}{2} \left(\frac{m}{2}\right) RT \]

For helium:
\[ KE_{He} = \frac{3}{2} \left(\frac{m}{4}\right) RT \]

Now, we can find the ratio of kinetic energies:
\[ \frac{KE_{H_2}}{KE_{He}} = \frac{\frac{5}{2} \left(\frac{m}{2}\right) RT}{\frac{3}{2} \left(\frac{m}{4}\right) RT} \]

Canceling out common terms:
\[ = \frac{5 \cdot 4}{3 \cdot 2} = \frac{20}{6} = \frac{10}{3} \]

### Final Answers
1. The ratio of pressure \( P_{H_2} : P_{He} = 2 : 1 \)
2. The ratio of total kinetic energy \( KE_{H_2} : KE_{He} = 10 : 3 \)