Two flasks A and B have equal volumes. A is maintained at 300 K and B at 600 K. A contains `H_(2)` gas, B has an equal mass of `CO_(2)` gas. Find the ratio of total K.E. of gases in flask A to that of B.

Options

A. `1:2`
B. `11:1`
C. `33:2`
D. `55:7`

Explanation

To solve the problem, we need to calculate the total kinetic energy of the gases in both flasks A and B and then find the ratio of these energies.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Flask A contains \( H_2 \) gas at \( T_A = 300 \, K \).
- Flask B contains \( CO_2 \) gas at \( T_B = 600 \, K \).
- Both flasks have equal volumes and contain equal masses of gas.

2. **Calculate the Moles of Each Gas:**
- Let the mass of gas in each flask be \( m \).
- For \( H_2 \):
\[ n_A = \frac{m}{M_{H_2}} = \frac{m}{2 \, g/mol} \]
- For \( CO_2 \):
\[ n_B = \frac{m}{M_{CO_2}} = \frac{m}{44 \, g/mol} \]

3. **Calculate the Total Kinetic Energy for Each Flask:**
- The formula for total kinetic energy (K.E.) of an ideal gas is given by:
\[ KE = \frac{3}{2} nRT \]
- For flask A (containing \( H_2 \)):
\[ KE_A = \frac{3}{2} n_A R T_A = \frac{3}{2} \left(\frac{m}{2}\right) R (300) \]
- For flask B (containing \( CO_2 \)):
\[ KE_B = \frac{3}{2} n_B R T_B = \frac{3}{2} \left(\frac{m}{44}\right) R (600) \]

4. **Substituting the Values:**
- Substitute \( n_A \) and \( n_B \) into the kinetic energy equations:
\[ KE_A = \frac{3}{2} \left(\frac{m}{2}\right) R (300) = \frac{3mR \cdot 300}{4} \]
\[ KE_B = \frac{3}{2} \left(\frac{m}{44}\right) R (600) = \frac{3mR \cdot 600}{88} \]

5. **Simplifying the Kinetic Energies:**
- Simplifying \( KE_A \):
\[ KE_A = \frac{3mR \cdot 300}{4} = \frac{900mR}{4} = 225mR \]
- Simplifying \( KE_B \):
\[ KE_B = \frac{3mR \cdot 600}{88} = \frac{1800mR}{88} = \frac{225mR}{11} \]

6. **Finding the Ratio of Total K.E.:**
- Now, we find the ratio \( \frac{KE_A}{KE_B} \):
\[ \frac{KE_A}{KE_B} = \frac{225mR}{\frac{225mR}{11}} = 11 \]

### Final Answer:
The ratio of total kinetic energy of gases in flask A to that of B is:
\[ \text{Ratio} = 11 : 1 \]