The mean free path of molecules of a certain gas at STP is 1500d, where d is the diameter of the gas molecules . While maintaining the standard pressure, the mean free path of the molecules at 373K is approximtely:

Options

A. `2049d`
B. `750d`
C. `1500d`
D. `1098d`

Answer

To solve the problem of finding the mean free path of gas molecules at 373K, we can follow these steps:

### Step 1: Understand the relationship between mean free path and temperature
The mean free path (\( \lambda \)) of gas molecules is given by the formula:
\[ \lambda = \frac{d^2}{\sqrt{2} \cdot n \cdot \sigma} \]
where \( n \) is the number density of the gas molecules and \( \sigma \) is the collision cross-section. However, for our purposes, we can use the simplified relationship that states:
\[ \lambda \propto \frac{T}{P} \]
Since the pressure \( P \) is constant, we can say:
\[ \frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1} \]

### Step 2: Identify the given values
From the problem, we have:
- Mean free path at STP (\( \lambda_1 \)) = \( 1500d \)
- Temperature at STP (\( T_1 \)) = 273 K
- Temperature at 373 K (\( T_2 \)) = 373 K

### Step 3: Set up the equation using the relationship
Using the relationship derived in Step 1:
\[ \frac{\lambda_2}{1500d} = \frac{373}{273} \]

### Step 4: Solve for \( \lambda_2 \)
Now, we can rearrange the equation to find \( \lambda_2 \):
\[ \lambda_2 = 1500d \cdot \frac{373}{273} \]

### Step 5: Calculate the numerical value
Now, we need to compute:
\[ \lambda_2 = 1500d \cdot \frac{373}{273} \approx 1500d \cdot 1.366 \]
Calculating this gives:
\[ \lambda_2 \approx 2049d \]

### Step 6: Conclusion
Thus, the mean free path of the molecules at 373K is approximately:
\[ \lambda_2 \approx 2049d \]

### Final Answer
The correct option is \( 2049d \).

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