In two vessels of 1 litre each at athe same temperature 1g of `H_(2)` and 1g of `CH_(4)` are taken. For these gases:

Options

A. `V_("rms")` values will be same
B. Kinetic energy per mol will be same
C. Total kinetic energy will same
D. Pressure will be same

Explanation

To solve the problem, we need to analyze the situation involving two gases, hydrogen (H₂) and methane (CH₄), in separate 1-liter vessels at the same temperature. We will calculate the root mean square velocity (VRMS), kinetic energy per mole, total kinetic energy, and pressure for both gases.

### Step 1: Calculate the Molar Mass of Each Gas
- **Molar Mass of H₂**:
- H has an atomic mass of approximately 1 g/mol, so for H₂:
\[ \text{Molar Mass of H₂} = 2 \times 1 = 2 \text{ g/mol} \]

- **Molar Mass of CH₄**:
- C has an atomic mass of approximately 12 g/mol and H has an atomic mass of approximately 1 g/mol. Thus:
\[ \text{Molar Mass of CH₄} = 12 + 4 \times 1 = 16 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of Each Gas
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

- For H₂:
\[ n_{H₂} = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ moles} \]

- For CH₄:
\[ n_{CH₄} = \frac{1 \text{ g}}{16 \text{ g/mol}} = 0.0625 \text{ moles} \]

### Step 3: Calculate the Root Mean Square Velocity (VRMS)
The formula for VRMS is:
\[ V_{RMS} = \sqrt{\frac{3RT}{M}} \]
Where:
- R = 8.314 J/(mol·K)
- T = Temperature in Kelvin
- M = Molar mass in kg/mol (convert g/mol to kg/mol by dividing by 1000)

- For H₂:
\[ V_{RMS(H₂)} = \sqrt{\frac{3RT}{0.002}} \quad \text{(since 2 g/mol = 0.002 kg/mol)} \]

- For CH₄:
\[ V_{RMS(CH₄)} = \sqrt{\frac{3RT}{0.016}} \quad \text{(since 16 g/mol = 0.016 kg/mol)} \]

### Step 4: Compare VRMS Values
Since the temperature (T) and R are the same for both gases, we can compare:
\[ \frac{V_{RMS(H₂)}}{V_{RMS(CH₄)}} = \sqrt{\frac{0.016}{0.002}} = \sqrt{8} = 2\sqrt{2} \]
This shows that VRMS for H₂ is greater than that for CH₄.

### Step 5: Calculate Kinetic Energy Per Mole
The kinetic energy per mole is given by:
\[ KE = \frac{3}{2} RT \]
Since R and T are the same for both gases, the kinetic energy per mole will also be the same for H₂ and CH₄.

### Step 6: Calculate Total Kinetic Energy
Total kinetic energy is given by:
\[ KE_{total} = n \cdot KE \]
- For H₂:
\[ KE_{total(H₂)} = 0.5 \cdot \frac{3}{2} RT \]

- For CH₄:
\[ KE_{total(CH₄)} = 0.0625 \cdot \frac{3}{2} RT \]

Since the number of moles is different, the total kinetic energy will also be different.

### Step 7: Calculate Pressure
Using the ideal gas law:
\[ PV = nRT \]
Since V and T are constant, pressure is directly proportional to the number of moles:
\[ P_{H₂} \propto n_{H₂} \quad \text{and} \quad P_{CH₄} \propto n_{CH₄} \]
Thus, the pressures will be different due to the difference in moles.

### Summary of Findings
1. **VRMS**: H₂ has a higher VRMS than CH₄.
2. **Kinetic Energy per Mole**: Same for both gases.
3. **Total Kinetic Energy**: Different for H₂ and CH₄ due to different moles.
4. **Pressure**: Different for both gases due to different moles.