By how many folds the temperature of a gas would increase when the r.m.s. velocity of gas molecules in a closed container of fixed volume is increased from `5 xx 10^(4) cm s^(-1) " to " 10 xx 10^(4) cm s^(-1)`?

Options

A. 0.5 times
B. 2 times
C. 4 times
D. 16 times

Explanation

To solve the problem, we need to determine how many folds the temperature of a gas increases when the root mean square (r.m.s.) velocity of gas molecules is increased from \(5 \times 10^4 \, \text{cm/s}\) to \(10 \times 10^4 \, \text{cm/s}\).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial r.m.s. velocity, \(v_1 = 5 \times 10^4 \, \text{cm/s}\)
- Final r.m.s. velocity, \(v_2 = 10 \times 10^4 \, \text{cm/s}\)

2. **Use the Relationship Between r.m.s. Velocity and Temperature**:
The r.m.s. velocity (\(v_{\text{rms}}\)) of gas molecules is related to the temperature (\(T\)) by the equation:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the gas constant and \(M\) is the molar mass of the gas. From this equation, we can see that:
\[ v_{\text{rms}} \propto \sqrt{T} \]
This implies:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]

3. **Set Up the Equation**:
Substituting the values of \(v_1\) and \(v_2\):
\[ \frac{5 \times 10^4}{10 \times 10^4} = \sqrt{\frac{T_1}{T_2}} \]
Simplifying the left side:
\[ \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

4. **Square Both Sides**:
To eliminate the square root, square both sides of the equation:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
This gives:
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

5. **Rearranging the Equation**:
Rearranging the equation to find \(T_2\):
\[ T_2 = 4T_1 \]

6. **Conclusion**:
The temperature \(T_2\) is 4 times the initial temperature \(T_1\). Thus, the temperature of the gas increases by a factor of 4.

### Final Answer:
The temperature of the gas increases by **4 folds**.