Question

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27°C is

• A. 327°C
• B. 627°C
• C. 927°C
• D. 1227°C

Answer 1

The Correct Option is A

The problem involves determining the temperature at which the kinetic energy of oxygen molecules is double that of its kinetic energy at 27°C. To solve this, we use the relationship between the average kinetic energy of gas molecules and temperature.

The average kinetic energy of gas molecules is given by the formula:

E = \frac{3}{2}kT

Where:

• E is the average kinetic energy.
• k is the Boltzmann constant.
• T is the absolute temperature in Kelvin.

To find the temperature at which the kinetic energy is double that at a certain temperature, we set up the relationship:

2E_{1} = E_{2}

Where E_1 is the initial kinetic energy at 27°C, and E_2 is the new kinetic energy when doubled. Substituting the energy formulas, we have:

2 \left(\frac{3}{2}kT_1\right) = \frac{3}{2}kT_2

Canceling common factors and solving for T_2 gives:

T_2 = 2T_1

Convert the initial temperature from Celsius to Kelvin:

T_1 = 27 + 273.15 = 300.15 \, \text{K}

Substitute and calculate T_2:

T_2 = 2 \times 300.15 = 600.3 \, \text{K}

Convert back from Kelvin to Celsius:

T_2 = 600.3 - 273.15 = 327.15 \approx 327^\circ\text{C}

Therefore, the correct answer is 327°C.