The temperature of low pressure hydrogen is reduced from `100^@ C `to `20^@ C`. The rms speed of its molecules decreases by, approximately

Options

A. 0.89
B. `79%`
C. `46%`
D. `11.3%`

Answer

To solve the problem of how the root mean square (rms) speed of hydrogen molecules decreases when the temperature is reduced from 100°C to 20°C, we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
The first step is to convert the given temperatures from degrees Celsius to Kelvin. The conversion formula is:
\[ T(K) = T(°C) + 273 \]

- For \( T_1 = 100°C \):
\[ T_1 = 100 + 273 = 373 \, K \]

- For \( T_2 = 20°C \):
\[ T_2 = 20 + 273 = 293 \, K \]

### Step 2: Understand the relationship between rms speed and temperature
The root mean square speed \( V_{rms} \) of gas molecules is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
Where:
- \( R \) is the gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molecular mass of the gas.

From this formula, we can see that \( V_{rms} \) is directly proportional to the square root of the temperature:
\[ V_{rms} \propto \sqrt{T} \]

### Step 3: Calculate the initial and final rms speeds
Using the proportionality, we can express the change in rms speed as:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{T_2}{T_1}} \]

Substituting the values:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{293}{373}} \]

### Step 4: Calculate the ratio
Now we need to compute the numerical value:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{0.785} \approx 0.886 \]

### Step 5: Calculate the percentage decrease
To find the percentage decrease in rms speed, we can use the formula:
\[ \text{Percentage Decrease} = \left(1 - \frac{V_{rms2}}{V_{rms1}}\right) \times 100 \]

Substituting the calculated ratio:
\[ \text{Percentage Decrease} = \left(1 - 0.886\right) \times 100 \approx 11.4\% \]

### Conclusion
The rms speed of hydrogen molecules decreases by approximately **11.4%** when the temperature is reduced from 100°C to 20°C.

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