JEE Mains · Physics · STD 11 - 11. thermodynamics
If one mole of an ideal gas at \(\left( P _{1}, V _{1}\right)\) is allowed to expand reversibly and isothermally (\(A\) to \(B\) ) its pressure is reduced to one-half of the original pressure (see \(figure\)). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value \(( B \rightarrow C ) .\) Then it is restored to its initial state by a reversible adiabatic compression (\(C\) to \(A\)). The net workdone by the gas is equal to ...... .
- A \(RT \left(\ln 2-\frac{1}{2(\gamma-1)}\right)\)
- B \(-\frac{ RT }{2(\gamma-1)}\)
- C \(0\)
- D \(RT \ln 2\)
Answer & Solution
Correct Answer
(A) \(RT \left(\ln 2-\frac{1}{2(\gamma-1)}\right)\)
Step-by-step Solution
Detailed explanation
\(A-B=\) isothermal process \(W _{ AB }= P _{1} V _{1} \ln \left[\frac{2 V _{1}}{ V _{1}}\right]= P _{1} V _{1} \ln (2)\) \(B - C \rightarrow\) Isochoric process \(W _{ BC }=0\) \(C - A \rightarrow\) Adiabatic process…
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