If the volume of a gas is doubled at constant pressure, the average translational kinetic energy of its molecules will

Options

A. be doubled
B. remain the same
C. increase by a factor
D. become fore times

Explanation

To solve the problem, we need to understand the relationship between the volume of a gas, its pressure, and the average translational kinetic energy of its molecules. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Analyze the Given Conditions
We are told that the volume of the gas is doubled at constant pressure. This means:
- Initial volume = \( V \)
- Final volume = \( 2V \)
- Pressure remains constant.

### Step 3: Determine the Effect on Temperature
Since pressure is constant and the volume is doubled, we can use the ideal gas law to find the relationship between temperature and volume. Rearranging the ideal gas law gives us:
\[ T = \frac{PV}{nR} \]
If we denote the initial temperature as \( T_1 \) and the final temperature as \( T_2 \), we can express the initial and final states as:
- Initial state: \( T_1 = \frac{PV}{nR} \)
- Final state: \( T_2 = \frac{P(2V)}{nR} = \frac{2PV}{nR} = 2T_1 \)

Thus, when the volume is doubled at constant pressure, the temperature also doubles.

### Step 4: Relate Temperature to Average Translational Kinetic Energy
The average translational kinetic energy (\( U \)) of the gas molecules is given by the equation:
\[ U = \frac{3}{2} nRT \]
From this equation, we can see that the average translational kinetic energy is directly proportional to the temperature.

### Step 5: Calculate the Change in Average Translational Kinetic Energy
Since we have established that the temperature doubles:
- Initial average translational kinetic energy: \( U_1 = \frac{3}{2} nRT_1 \)
- Final average translational kinetic energy: \( U_2 = \frac{3}{2} nRT_2 = \frac{3}{2} nR(2T_1) = 2 \left( \frac{3}{2} nRT_1 \right) = 2U_1 \)

This shows that the average translational kinetic energy also doubles.

### Conclusion
Thus, if the volume of a gas is doubled at constant pressure, the average translational kinetic energy of its molecules will also double.

### Final Answer
The average translational kinetic energy of its molecules will double.

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