What is r.m.s speed of `O_(2)` molecule if its kinetic energy is 2 k cal `"mol"^(-1)`?

Options

A. `7.24 xx 10^(2)` m/sec
B. `3.5 xx 10^(2)` m/sec
C. `2 xx 10^(1)` m/sec
D. `3.5 xx 10^(4)` m/sec

Answer

To find the root mean square (r.m.s) speed of an O₂ molecule given its kinetic energy, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and r.m.s speed
The kinetic energy (KE) of a gas can be expressed as:
\[ KE = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

The r.m.s speed (\( u_{rms} \)) is given by:
\[ u_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( M \) = molar mass of the gas in kg/mol

### Step 2: Relate kinetic energy to r.m.s speed
From the kinetic energy formula, we can express \( RT \) in terms of kinetic energy:
\[ RT = \frac{2KE}{3} \]

### Step 3: Substitute \( RT \) into the r.m.s speed formula
Substituting \( RT \) into the r.m.s speed formula gives:
\[ u_{rms} = \sqrt{\frac{3 \cdot \frac{2KE}{3}}{M}} \]
This simplifies to:
\[ u_{rms} = \sqrt{\frac{2KE}{M}} \]

### Step 4: Convert kinetic energy from kcal/mol to joules
Given that the kinetic energy is 2 kcal/mol, we need to convert this to joules:
\[ 1 \text{ kcal} = 4184 \text{ J} \]
Thus,
\[ KE = 2 \text{ kcal/mol} \times 4184 \text{ J/kcal} = 8368 \text{ J/mol} \]

### Step 5: Find the molar mass of O₂
The molar mass of O₂ is:
\[ M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol} \]

### Step 6: Substitute values into the r.m.s speed formula
Now we can substitute the values of \( KE \) and \( M \) into the r.m.s speed formula:
\[ u_{rms} = \sqrt{\frac{2 \times 8368 \text{ J/mol}}{32 \times 10^{-3} \text{ kg/mol}}} \]

### Step 7: Calculate \( u_{rms} \)
Calculating the above expression:
\[ u_{rms} = \sqrt{\frac{16736}{0.032}} \]
\[ u_{rms} = \sqrt{523000} \]
\[ u_{rms} \approx 724.08 \text{ m/s} \]

### Final Answer
The r.m.s speed of the O₂ molecule is approximately:
\[ \boxed{724.08 \text{ m/s}} \]