Two vessels A and B having equal volume contain equal masses of hydrogen in A and helium in B at 300 K. Then, mark the correct statement?

Options

A. The pressure exerted by hydrogen is half that exerted by helium
B. The pressure exerted by hydrogen is equal to that exacted by helium
C. Average KE of the molecule of hydrogen is half the average KE of the molecules of helium
D. The pressure exeted by hydrogen is twice theat exerted by helium

Explanation

To solve the problem, we need to analyze the given information about the two vessels containing equal masses of hydrogen and helium at the same temperature and volume. We will derive the relationships for pressure and average kinetic energy of the gases.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Vessels A and B have equal volumes.
- Equal masses of hydrogen (H₂) in vessel A and helium (He) in vessel B.
- Temperature (T) = 300 K.

2. **Use the Ideal Gas Law:**
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \(P\) = pressure,
- \(V\) = volume,
- \(n\) = number of moles,
- \(R\) = universal gas constant,
- \(T\) = temperature in Kelvin.

3. **Calculate the Number of Moles:**
The number of moles \(n\) can be calculated using the formula:
\[ n = \frac{m}{M} \]
where:
- \(m\) = mass of the gas,
- \(M\) = molar mass of the gas.

For hydrogen (H₂), the molar mass \(M_H = 2 \, \text{g/mol}\).
For helium (He), the molar mass \(M_{He} = 4 \, \text{g/mol}\).

4. **Calculate Moles for Each Gas:**
Since both vessels contain equal masses:
\[ n_H = \frac{m}{2} \quad \text{(for hydrogen)} \]
\[ n_{He} = \frac{m}{4} \quad \text{(for helium)} \]

5. **Express Pressure for Each Gas:**
Using the ideal gas law for both gases:
\[ P_H V = n_H R T \quad \Rightarrow \quad P_H = \frac{n_H R T}{V} \]
\[ P_{He} V = n_{He} R T \quad \Rightarrow \quad P_{He} = \frac{n_{He} R T}{V} \]

6. **Substituting the Number of Moles:**
Substitute \(n_H\) and \(n_{He}\) into the pressure equations:
\[ P_H = \frac{\left(\frac{m}{2}\right) R T}{V} = \frac{m R T}{2V} \]
\[ P_{He} = \frac{\left(\frac{m}{4}\right) R T}{V} = \frac{m R T}{4V} \]

7. **Finding the Pressure Ratio:**
Now, we can find the ratio of pressures:
\[ \frac{P_H}{P_{He}} = \frac{\frac{m R T}{2V}}{\frac{m R T}{4V}} = \frac{4}{2} = 2 \]
This means:
\[ P_H = 2 P_{He} \]

8. **Average Kinetic Energy:**
The average kinetic energy of a gas molecule is given by:
\[ KE = \frac{3}{2} k T \]
where \(k\) is the Boltzmann constant. Since both gases are at the same temperature (300 K), the average kinetic energy for both gases will be equal:
\[ KE_H = KE_{He} \]

9. **Conclusion:**
- The pressure exerted by hydrogen is twice that exerted by helium.
- The average kinetic energy of the molecules of hydrogen is equal to that of helium.

### Correct Statement:
The correct statement is that the pressure exerted by hydrogen is twice the pressure exerted by helium.