Two flasks X and Y have capacity 1 L and 2 L respectively and each of them contains 1 mole of a gas. The temperature of the flasks are so adjusted that average speed of molecules in X is twice as those in Y. The pressure in flask X would be

Options

A. same as that in Y
B. half of that in Y
C. twice of that in Y
D. 8 times of that in Y.

Explanation

To solve the problem, we need to determine the pressure in flask X given the conditions described. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- Flask X has a volume of 1 L and contains 1 mole of gas.
- Flask Y has a volume of 2 L and also contains 1 mole of gas.
- The average speed of gas molecules in flask X (U_x) is twice that of flask Y (U_y).

### Step 2: Use the Kinetic Theory of Gases
According to the kinetic theory of gases, the pressure (P) of an ideal gas can be expressed as:
\[ P = \frac{1}{3} \frac{n \cdot m \cdot U^2}{V} \]
Where:
- P = pressure
- n = number of molecules
- m = mass of one molecule
- U = average speed of the molecules
- V = volume of the gas

### Step 3: Set Up the Equations for Each Flask
For flask X:
\[ P_X = \frac{1}{3} \frac{n_X \cdot m \cdot U_X^2}{V_X} \]
For flask Y:
\[ P_Y = \frac{1}{3} \frac{n_Y \cdot m \cdot U_Y^2}{V_Y} \]

### Step 4: Analyze the Number of Molecules
Since both flasks contain 1 mole of gas, the number of molecules in each flask is the same:
\[ n_X = n_Y \]

### Step 5: Substitute the Average Speeds
Given that \( U_X = 2U_Y \), we can substitute this into the equation for flask X:
\[ P_X = \frac{1}{3} \frac{n_X \cdot m \cdot (2U_Y)^2}{V_X} \]
\[ P_X = \frac{1}{3} \frac{n_X \cdot m \cdot 4U_Y^2}{V_X} \]

### Step 6: Relate the Volumes
The volume of flask X is \( V_X = 1 \, \text{L} \) and the volume of flask Y is \( V_Y = 2 \, \text{L} \).

### Step 7: Write the Equation for Flask Y
For flask Y:
\[ P_Y = \frac{1}{3} \frac{n_Y \cdot m \cdot U_Y^2}{V_Y} \]
Substituting \( n_Y = n_X \) and \( V_Y = 2 \):
\[ P_Y = \frac{1}{3} \frac{n_X \cdot m \cdot U_Y^2}{2} \]

### Step 8: Find the Ratio of Pressures
Now, we can find the ratio of the pressures:
\[ \frac{P_X}{P_Y} = \frac{\frac{1}{3} \frac{n_X \cdot m \cdot 4U_Y^2}{V_X}}{\frac{1}{3} \frac{n_X \cdot m \cdot U_Y^2}{2}} \]

### Step 9: Simplify the Ratio
This simplifies to:
\[ \frac{P_X}{P_Y} = \frac{4U_Y^2/V_X}{U_Y^2/(2)} \]
\[ \frac{P_X}{P_Y} = \frac{4 \cdot 2}{1} = 8 \]

### Step 10: Conclusion
Thus, we find that:
\[ P_X = 8 \cdot P_Y \]

### Final Answer
The pressure in flask X is 8 times that in flask Y.

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