Question

In a process carried out at constant volume, the heat absorbed (Q\(_{\text{v}}\)) is equal to the change in:

• A. Gibbs free energy
• B. Enthalpy
• C. Internal energy
• D. Work done

Hint:

Remember these thermodynamic definitions:
- Constant Volume: \( Q_v = \Delta U \) (Internal Energy).
- Constant Pressure: \( Q_p = \Delta H \) (Enthalpy).
This is a core concept tested frequently in chemical and materials thermodynamics.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks to identify which state function is equal to the heat absorbed (\( Q_v \)) during a thermodynamic process carried out at constant volume.

Step 2: Key Formula or Approach:
By the First Law of Thermodynamics, the change in internal energy (\( dU \)) of a system is given by:
\[ dU = dQ - dW \]
where:
\( dQ \) is the heat added to the system.
\( dW \) is the work done by the system.
The boundary work for a simple compressible system is given by:
\[ dW = P dV \]

Step 3: Detailed Explanation:

• Constant Volume Condition:
- For an isochoric (constant volume) process, the change in volume is zero:
\[ dV = 0 \]
- Substituting this into the work equation shows that no boundary work is performed:
\[ dW = P(0) = 0 \]
- Substituting this back into the First Law:
\[ dU = dQ_v - 0 \implies dQ_v = dU \]
- Thus, all heat energy absorbed by the system at constant volume goes entirely into changing the internal energy of the system.

• Comparison with Constant Pressure:
- For a process at constant pressure, the heat absorbed \( dQ_p \) equals the change in Enthalpy (\( dH \)), because:
\[ dH = dU + d(PV) = dQ_p \]


Step 4: Final Answer:
The heat absorbed at constant volume is equal to the change in internal energy.
Therefore, the correct choice is option (C).