Question

What happens when an ideal gas undergoes isothermal expansion into vacuum?

• A. \(w=0,\;Q=0\)
• B. \(Q=+ve,\;w=-ve\)
• C. \(w=-ve,\;Q=+ve\)
• D. \(w=0\)

Hint:

For free expansion of an ideal gas: \[ \boxed{ \begin{aligned} P_{\text{ext}}&=0\\ w&=0\\ \Delta U&=0\;(\text{isothermal})\\ Q&=0 \end{aligned} } \] Always remember: \[ \boxed{\Delta U=Q+w} \] is the First Law of Thermodynamics.

Answer 1

The Correct Option is A

Concept: Expansion of a gas into a vacuum is called

free expansion. For free expansion, \[ P_{\text{ext}}=0. \] The work done is \[ w=-P_{\text{ext}}\Delta V. \] Since the external pressure is zero, \[ \boxed{w=0.} \] For an ideal gas, \[ \Delta U=nC_V\Delta T. \] In an isothermal process, \[ \Delta T=0, \] therefore, \[ \boxed{\Delta U=0.} \] Using the first law of thermodynamics, \[ \Delta U=Q+w. \] Since both \(\Delta U\) and \(w\) are zero, \[ \boxed{Q=0.} \]

Step 1: Calculate the work done.
For free expansion, \[ P_{\text{ext}}=0. \] Hence, \[ w=-P_{\text{ext}}\Delta V=0. \] Thus, \[ \boxed{w=0.} \]

Step 2: Determine the change in internal energy.
Since the expansion is isothermal, \[ \Delta T=0. \] For an ideal gas, \[ \boxed{\Delta U=0.} \]

Step 3: Apply the First Law of Thermodynamics.
Using, \[ \Delta U=Q+w, \] we get \[ 0=Q+0. \] Therefore, \[ \boxed{Q=0.} \] Hence, \[ \boxed{ w=0,\qquad Q=0. } \] Therefore, \[ \boxed{\textbf{Option (A)}} \] is the correct answer.