Q. An ideal gas is allowed to expand freely against a vacuum in a rigid insulated container. The gas undergoes
A) An increase in its internal energy
B) A decrease in its internal energy
C) Neither an increase nor a decrease in its temperature or internal energy
D) A decrease in temperature
Explanation
(c) For vacuum, pressure p=0
Hence, work done $= p\mathrm{\Delta}V = 0$
According to first law of thermodynamics
$Q = \mathrm{\Delta}U + p\mathrm{\Delta}V$
$\ \therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q = \mathrm{\Delta}U$
Hence the gas undergoes neither an increase nor a decrease in its temperature or internal energy.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.