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
Correct Answer: C
Solution :
For vacuum, pressure \[p=0\] Hence, work done \[=p\Delta V=0\] According to first law of thermodynamics \[Q=\Delta U+p\Delta V\] \[\therefore \] \[Q=\Delta U\] Hence the gas undergoes neither an increase nor a. decrease in its temperature or internal energy.
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