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What happens in a reversible adiabatic expansion process?
Answer & Solution
Correct Answer:
Option
B
Since for an adiabatic process \[T{V^{\gamma - 1}} = {\rm{CONSTANT}}\] and on integration of this equation we get \[ln\left( {\frac{{{T_2}}}{{{T_1}}}} \right) = - \left( {\left( {\gamma - 1} \right)ln\frac{{{V_2}}}{{{V_1}}}} \right)\] since for expansion process \[{V_2} > {V_1}\] Hence \[ \Rightarrow {T_2} < T_1\]
So, cooling takes place.
So, cooling takes place.
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