JEE Mains · Physics · STD 11 - 11. thermodynamics
A certain amount of gas of volume \(V\) at \(27^{o}\,C\) temperature and pressure \(2 \times 10^{7} \;Nm ^{-2}\) expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use \(\gamma=1.5\) )
- A \(3.536 \times 10^{5}\,Pa\)
- B \(3.536 \times 10^{6}\,Pa\)
- C \(1.25 \times 10^{6}\,Pa\)
- D \(1.25 \times 10^{5}\,Pa\)
Answer & Solution
Correct Answer
(B) \(3.536 \times 10^{6}\,Pa\)
Step-by-step Solution
Detailed explanation
\(P _{1}=2 \times 10^{7} Pa\) \(P _{1} V _{1}= P _{2} V _{2}\) Since \(V_{2}=2 V_{1}\) Hence \(P_{2}=P_{1} / 2\) (isothermal expansion) \(P _{2}=1 \times 10^{7} Pa\) \(P _{2}\left( V _{2}\right)^{\gamma}= P _{3}\left(2 V _{2}\right)^{\gamma}\)…
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