Free expansion (Joule expansion) of an ideal gas: A rigid, perfectly insulated container of volume V is split into two equal halves. One side is vacuum; the other contains 1 mol ideal gas at 298 K. After the partition is broken, what is the final temperature of the gas?
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AGreater than 298 K
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B298 K
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CLess than 298 K
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DCannot be determined
Answer
Correct Answer: 298 K
Explanation
Introduction / Context:Free expansion into vacuum (Joule expansion) is a classic thermodynamics scenario used to distinguish ideal from real gas behavior. It illustrates how internal energy depends on temperature for an ideal gas.
Given Data / Assumptions:
- Rigid container (no boundary work).
- Perfectly insulated (adiabatic: Q = 0).
- Ideal gas with temperature-independent heat capacity.
- Initial T = 298 K; final volume doubles after partition removal.
Concept / Approach:First law for a closed, adiabatic, rigid system: ΔU = Q − W = 0 − 0 = 0. For an ideal gas, U depends only on T. Therefore ΔT = 0 and the final temperature equals the initial temperature.
Step-by-Step Solution:
Apply ΔU = n * Cv * (T2 − T1) = 0.Since n and Cv are positive, T2 − T1 = 0.Hence T2 = 298 K.Verification / Alternative check:Real gases may show small Joule effects due to non-ideal interactions, but the ideal-gas model assumes zero; thus, temperature remains unchanged.
Why Other Options Are Wrong:
- Greater/Less than 298 K: Would require net heat or work interactions, absent here, or non-ideal effects not assumed.
- Cannot be determined: It can, from the first law and ideal-gas property of U(T) only.
Common Pitfalls:Confusing free expansion with throttling (Joule–Thomson) where enthalpy, not internal energy, is constant and real-gas effects matter.
Final Answer:298 K