Question

The translational kinetic energy of the molecules of a gas at absolute temperature ($T$) can be doubled

• A. by increasing $T$ to $4T$
• B. by increasing $T$ to $2T$
• C. by decreasing $T$ to $\frac{T}{2}$
• D. by increasing $T$ to $\sqrt{2}T$

Hint:

Do not confuse kinetic energy with root-mean-square velocity ($v_{\text{rms}}$)! The velocity tracks with the square root of temperature ($v_{\text{rms}} \propto \sqrt{T}$), meaning doubling the velocity would require a $4T$ temperature increase. However, since kinetic energy scales directly and linearly with temperature ($E \propto T$), a simple doubling of kinetic energy requires exactly a doubling of temperature ($2T$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the required change in the absolute temperature of an ideal gas system such that the mean translational kinetic energy of its constituent molecules is doubled.

Step 2: Key Formula or Approach:
According to the Kinetic Theory of Gases, the average translational kinetic energy ($E$) of a gas molecule depends exclusively on its absolute temperature $T$ through the relation: $$E = \frac{3}{2} k_B T$$ Where $k_B$ is the Boltzmann constant. This establishes a direct mathematical proportionality: $$E \propto T$$

Step 3: Detailed Explanation:
Let the initial kinetic energy at temperature $T_1 = T$ be $E_1$. Let the final kinetic energy at temperature $T_2$ be $E_2$. From our direct proportionality relationship, we can set up a ratio equation: $$\frac{E_2}{E_1} = \frac{T_2}{T_1}$$ We want to double the translational kinetic energy, meaning $E_2 = 2E_1$: $$\frac{2E_1}{E_1} = \frac{T_2}{T} \implies 2 = \frac{T_2}{T}$$ Isolating the final temperature parameter $T_2$: $$T_2 = 2T$$ Therefore, the absolute temperature must be increased to exactly twice its initial value.

Step 4: Final Answer:
The translational kinetic energy can be doubled by increasing $T$ to $2T$, matching option (B).