Given below are two statement statement I :The temperature of gas is `-73^@C`. When the gas is heated to `527^@C`,the root mean square speed of the molecules is doubled statement II :The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules. inthe light of the above statement , choose the correct answer from the option given below:

Options

A. Statement I is false but Staement II is true
B. Both Statement I and Statement II are true
C. Both Statement I and Statement II are false
D. Statement I is true but Staement II is false

Answer

To solve the problem, we need to analyze the two statements provided.

### Step 1: Analyze Statement I
Statement I claims that when a gas is heated from -73°C to 527°C, the root mean square (RMS) speed of the gas molecules is doubled.

1. **Convert Temperatures to Kelvin**:
- Temperature \( T_1 = -73°C = 273 - 73 = 200 K \)
- Temperature \( T_2 = 527°C = 273 + 527 = 800 K \)

2. **RMS Speed Formula**:
The RMS speed of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas.

3. **Ratio of RMS Speeds**:
To find the relationship between the RMS speeds at the two temperatures, we can use:
\[ \frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2 \]
This shows that the RMS speed doubles when the temperature increases from 200 K to 800 K.

### Step 2: Analyze Statement II
Statement II states that the product of pressure and volume of an ideal gas is equal to the translational kinetic energy of the molecules.

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

2. **Translational Kinetic Energy**:
The translational kinetic energy (KE) of the gas molecules can be expressed as:
\[ KE = \frac{3}{2} nRT \]

3. **Equating the Two**:
From the ideal gas law, we can express \( PV \) as:
\[ PV = nRT \]
The translational kinetic energy is:
\[ KE = \frac{3}{2} PV \]
Thus, we can see that \( PV \) is indeed related to the translational kinetic energy, but it is not equal to it; rather, it is proportional to it.

### Conclusion
- **Statement I is True**: The RMS speed of the gas molecules is indeed doubled when the temperature is increased from -73°C to 527°C.
- **Statement II is False**: The product of pressure and volume is proportional to the translational kinetic energy but not equal to it.

### Final Answer
- Statement I is True.
- Statement II is False.