Question

The entropy change when 400 J of heat is absorbed reversibly at 400 K is:

• A. 0.5 J K$^{-1}$
• B. 1 J K$^{-1}$
• C. 2 J K$^{-1}$
• D. 4 J K$^{-1}$

Hint:

Ensure that the temperature is always in Kelvin (K) when calculating thermodynamic properties like entropy change.
If the temperature is given in Celsius, add 273.15 to convert it to Kelvin.
Pay attention to the sign: absorption of heat increases entropy ($+q$), while release of heat decreases entropy ($-q$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is from "Thermodynamics" (both in Physics and Chemistry) and asks us to find the entropy change ($\Delta S$) of a system when it absorbs a given amount of heat reversibly at a constant temperature.

Step 2: Key Formula or Approach:
The entropy change ($\Delta S$) of a thermodynamic system during a reversible isothermal process is given by the formula:
\[ \Delta S = \frac{q_{\text{rev}}}{T} \]
where $q_{\text{rev}}$ is the heat absorbed reversibly, and $T$ is the absolute temperature in Kelvin.

Step 3: Detailed Explanation:

• We are given:
Heat absorbed ($q_{\text{rev}}$) = $+400\text{ J}$ (the positive sign denotes that heat is absorbed by the system)
Absolute temperature ($T$) = $400\text{ K}$

• Substituting these values into the entropy change formula:
\[ \Delta S = \frac{400\text{ J}}{400\text{ K}} \]

• Performing the division:
\[ \Delta S = 1\text{ J K}^{-1} \]

• Entropy is a state function that measures the degree of randomness or disorder in the system.

• When a system absorbs heat, its molecular motion increases, causing a positive change in its entropy ($\Delta S \gt 0$).



Step 4: Final Answer:
The change in entropy is $1\text{ J K}^{-1}$, which corresponds to option (B).