When pressure remaining constant, at what temperature will the r.m.s. speed of a gas molecules increase by 10% of the r.m.s. speed at NTP?

Options

A. `57.3K`
B. `57.3^@ C`
C. `557.3K`
D. `-57.3^@ C`

Answer

To solve the problem of determining the temperature at which the root mean square (r.m.s.) speed of gas molecules increases by 10% at constant pressure, we can follow these steps:

### Step 1: Understand the r.m.s. speed formula
The r.m.s. speed \( V_{\text{rms}} \) of gas molecules is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Calculate the r.m.s. speed at NTP
At Normal Temperature and Pressure (NTP), the temperature \( T_0 \) is 273 K. Thus, the r.m.s. speed at NTP is:
\[ V_0 = \sqrt{\frac{3R \cdot 273}{M}} \]

### Step 3: Determine the increased r.m.s. speed
We need to find the new r.m.s. speed \( V_f \) after a 10% increase:
\[ V_f = V_0 + 0.1 V_0 = 1.1 V_0 \]

### Step 4: Set up the equation for the new r.m.s. speed
Using the r.m.s. speed formula for the new temperature \( T \):
\[ V_f = \sqrt{\frac{3RT}{M}} \]

### Step 5: Equate the two expressions for \( V_f \)
Now we can set the two expressions for \( V_f \) equal to each other:
\[ 1.1 V_0 = \sqrt{\frac{3RT}{M}} \]

### Step 6: Substitute \( V_0 \) into the equation
Substituting \( V_0 = \sqrt{\frac{3R \cdot 273}{M}} \) into the equation gives:
\[ 1.1 \sqrt{\frac{3R \cdot 273}{M}} = \sqrt{\frac{3RT}{M}} \]

### Step 7: Square both sides to eliminate the square root
Squaring both sides results in:
\[ (1.1)^2 \left(\frac{3R \cdot 273}{M}\right) = \frac{3RT}{M} \]
This simplifies to:
\[ 1.21 \cdot 3R \cdot 273 = 3RT \]

### Step 8: Cancel common terms and solve for \( T \)
We can cancel \( 3R \) from both sides (assuming \( R \neq 0 \)):
\[ 1.21 \cdot 273 = T \]

### Step 9: Calculate the final temperature
Now, calculating \( T \):
\[ T = 1.21 \cdot 273 \approx 330.03 \, \text{K} \]

### Final Answer
Thus, the temperature at which the r.m.s. speed of gas molecules increases by 10% at constant pressure is approximately:
\[ T \approx 330 \, \text{K} \]

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