Question

Which of the following statements is true? ($\Delta U = $ increase in internal energy, $dW = $ work done by the system)
\

• A. In an adiabatic process $\Delta U = dW$
  \
• B. In an adiabatic process $\Delta U = -dW$
  \
• C. In an isothermal process $\Delta U = -dW$
  \
• D. In an isothermal process $\Delta U = dW$

Hint:

Remember the physical meaning behind the negative sign! In an adiabatic process, if a gas does positive work ($dW > 0$) by expanding without an external heat source, it must consume its own internal energy stores to do so, causing temperature and internal energy to drop ($\Delta U < 0$). Hence, $\Delta U = -dW$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to identify the correct statement relating change in internal energy ($\Delta U$) and work done ($dW$) for basic thermodynamic processes according to the first law of thermodynamics.

Step 2: Key Formula or Approach:
The First Law of Thermodynamics is stated mathematically as: $$dQ = \Delta U + dW$$ Where $dQ$ is heat added, $\Delta U$ is the change in internal energy, and $dW$ is work done by the system. In an adiabatic process, no heat is exchanged with the surroundings: $dQ = 0$. In an isothermal process, temperature remains constant, which implies $\Delta U = 0$ for an ideal gas.

Step 3: Detailed Explanation:
Let's analyze the adiabatic condition first. Setting $dQ = 0$ in the first law equation yields: $$0 = \Delta U + dW \implies \Delta U = -dW$$ This matches option (B) exactly. For an isothermal process, because temperature is completely fixed, the internal energy does not change ($\Delta U = 0$). This reduces the first law to $dQ = dW$. It does not establish a general relation where $\Delta U = \pm dW$, which rules out options (C) and (D).

Step 4: Final Answer:
The correct relationship is given by option (B).