Question

The initial average kinetic energy of the molecules was E, when a gas sample is at 27$^\circ$C. When the gas is heated to 327$^\circ$C, then the final average kinetic energy will be ______.

• A. $\sqrt{2}E$
• B. 2E
• C. 300E
• D. 327E

Hint:

Always convert Celsius to Kelvin! A common trap is assuming that because $327$ isn't twice $27$, the energy doesn't double. Temperature formulas in physics almost universally demand the absolute thermodynamic scale.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A sample of gas is heated from a specific initial temperature to a specific final temperature. We need to determine how the average kinetic energy of its constituent molecules scales with this heating.

Step 2: Key Formula or Approach:
According to the Kinetic Theory of Gases, the average translational kinetic energy ($E_k$) of a gas molecule is directly proportional to the absolute temperature (measured strictly in Kelvin).
$$E_k = \frac{3}{2} k_B T$$
Therefore, $E_k \propto T$.
$$\frac{E_2}{E_1} = \frac{T_2}{T_1}$$

Step 3: Detailed Explanation:
1. Convert the initial temperature to Kelvin:
$$T_1 = 27^\circ\text{C} + 273 = 300 \text{ K}$$
At this temperature, the kinetic energy is $E_1 = E$.
2. Convert the final temperature to Kelvin:
$$T_2 = 327^\circ\text{C} + 273 = 600 \text{ K}$$
3. Set up the proportionality ratio:
$$\frac{E_2}{E_1} = \frac{T_2}{T_1}$$
$$\frac{E_2}{E} = \frac{600}{300}$$
$$\frac{E_2}{E} = 2$$
$$E_2 = 2E$$
Because the absolute temperature doubled (from 300K to 600K), the average kinetic energy perfectly doubles as well.

Step 4: Final Answer:
The final average kinetic energy is 2E, matching option (b).