The ratio of final to initial pressures of a gas when `u_(rms)` of a gas in a container is increased from `5 xx 10^(4) cm sec^(-1)" to " 10 xx 10^(4) cm sec^(-1)`

Explanation

To solve the problem of finding the ratio of final to initial pressures of a gas when the root mean square velocity (u_rms) is increased, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The root mean square velocity (u_rms) of a gas is given by the formula:
\[ u_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

2. **Use the Ideal Gas Law**: According to the ideal gas equation, we have:
\[ PV = nRT \]
From this, we can express \( RT \) as:
\[ RT = \frac{PV}{n} \]

3. **Substitute RT in u_rms Equation**: By substituting \( RT \) into the u_rms equation, we get:
\[ u_{rms} = \sqrt{\frac{3PV}{nM}} \]

4. **Identify Proportionality**: Since the volume \( V \), number of moles \( n \), and molar mass \( M \) are constants, we can conclude that:
\[ u_{rms} \propto \sqrt{P} \]
Therefore, we can write:
\[ \frac{u_{rms1}}{u_{rms2}} = \sqrt{\frac{P1}{P2}} \]

5. **Insert Given Values**: From the problem, we know:
- \( u_{rms1} = 5 \times 10^4 \, \text{cm/s} \)
- \( u_{rms2} = 10 \times 10^4 \, \text{cm/s} \)

Plugging these values into the equation gives:
\[ \frac{5 \times 10^4}{10 \times 10^4} = \sqrt{\frac{P1}{P2}} \]

6. **Simplify the Ratio**: Simplifying the left side:
\[ \frac{1}{2} = \sqrt{\frac{P1}{P2}} \]

7. **Square Both Sides**: To eliminate the square root, square both sides:
\[ \left(\frac{1}{2}\right)^2 = \frac{P1}{P2} \]
This results in:
\[ \frac{1}{4} = \frac{P1}{P2} \]

8. **Find the Final Ratio**: We need the ratio of final to initial pressure \( \frac{P2}{P1} \):
\[ \frac{P2}{P1} = 4 \]

### Final Answer:
The ratio of final to initial pressures \( P2 : P1 \) is \( 4 : 1 \).

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