The average translational K.E. in one millitre volume of oxygen at NTP is

Options

A. 0.15 J
B. 0.036 J
C. 0.56 J
D. 152 J

Answer

To solve the problem of finding the average translational kinetic energy in one milliliter volume of oxygen at NTP (Normal Temperature and Pressure), we can follow these steps:

### Step 1: Convert the Volume
Given that the volume is 1 milliliter, we need to convert this into cubic meters for standard SI units.
\[ 1 \text{ mL} = 1 \times 10^{-3} \text{ L} = 1 \times 10^{-6} \text{ m}^3 \]

### Step 2: Identify the Conditions at NTP
At Normal Temperature and Pressure (NTP), the temperature is 25 degrees Celsius, which we convert to Kelvin:
\[ T = 25 + 273 = 298 \text{ K} \]
The pressure at NTP is given as 1 atmosphere, which we convert to Pascals:
\[ P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \approx 10^5 \text{ Pa} \]

### Step 3: Determine the Degrees of Freedom
For diatomic gases like oxygen, the degrees of freedom (f) are:
- Translational degrees of freedom: 3
- Rotational degrees of freedom: 2
- Total degrees of freedom: 5

### Step 4: Calculate the Average Translational Kinetic Energy
The average translational kinetic energy (K.E.) can be calculated using the formula:
\[ \text{K.E.} = \frac{f}{2} k T \]
Where \( k \) is the Boltzmann constant, \( k \approx 1.38 \times 10^{-23} \text{ J/K} \).

However, we can also express the average translational kinetic energy in terms of pressure and volume:
\[ \text{K.E.} = \frac{3}{2} PV \]

### Step 5: Substitute the Values
Now we substitute the values of pressure and volume into the equation:
\[ \text{K.E.} = \frac{3}{2} \times (10^5 \text{ Pa}) \times (1 \times 10^{-6} \text{ m}^3) \]
Calculating this gives:
\[ \text{K.E.} = \frac{3}{2} \times 10^{-1} \text{ J} = 0.15 \text{ J} \]

### Conclusion
Thus, the average translational kinetic energy in one milliliter of oxygen at NTP is:
\[ \text{K.E.} = 0.15 \text{ J} \]

### Final Answer
The correct answer is \( 0.15 \text{ J} \).

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