Question

For an isothermal reversible compression of an ideal gas

• A. only \(\Delta E = 0\)
• B. only \(\Delta H = 0\)
• C. \(\Delta E = \Delta H = 0\)
• D. dQ = dE

Hint:

A key property of an ideal gas is that its internal energy and enthalpy depend *only* on temperature. Therefore, for *any* isothermal process (constant T) involving an ideal gas, you can immediately conclude that \(\Delta E = 0\) and \(\Delta H = 0\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to determine the changes in internal energy (\(\Delta E\) or \(\Delta U\)) and enthalpy (\(\Delta H\)) for an ideal gas undergoing an isothermal process (constant temperature).

Step 2: Key Formula or Approach:
For an ideal gas, the internal energy (\(E\)) is a function of temperature only. The change in internal energy is given by: \[ \Delta E = n C_v \Delta T \] where \(n\) is the number of moles, \(C_v\) is the molar heat capacity at constant volume, and \(\Delta T\) is the change in temperature.
Similarly, for an ideal gas, enthalpy (\(H\)) is also a function of temperature only. The change in enthalpy is given by: \[ \Delta H = n C_p \Delta T \] where \(C_p\) is the molar heat capacity at constant pressure.

Step 3: Detailed Explanation:
The process is described as "isothermal," which means the temperature remains constant throughout the process.
Therefore, the change in temperature is zero: \[ \Delta T = T_{final} - T_{initial} = 0 \] Now, let's calculate \(\Delta E\) and \(\Delta H\).
For the change in internal energy: \[ \Delta E = n C_v (0) = 0 \] For the change in enthalpy: \[ \Delta H = n C_p (0) = 0 \] Thus, for any isothermal process involving an ideal gas (be it compression, expansion, reversible, or irreversible), both the change in internal energy and the change in enthalpy are zero.
Let's analyze option (D): dQ = dE. From the first law of thermodynamics, \(dE = dQ + dW\). Since \(dE = 0\), we have \(dQ = -dW\). So, dQ is not equal to dE.

Step 4: Final Answer:
Since the process is isothermal (\(\Delta T = 0\)) and the substance is an ideal gas, both internal energy and enthalpy, being functions of temperature only, do not change. Therefore, \(\Delta E = 0\) and \(\Delta H = 0\).