JEE Mains · Physics · STD 11 - 11. thermodynamics
Starting at temperature \(300\; \mathrm{K},\) one mole of an ideal diatomic gas \((\gamma=1.4)\) is first compressed adiabatically from volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}=\frac{\mathrm{V}_{1}}{16} .\) It is then allowed to expand isobarically to volume \(2 \mathrm{V}_{2} \cdot\) If all the processes are the quasi-static then the final temperature of the gas (in \(\left. \mathrm{K}\right)\) is (to the nearest integer)
- A \(1818\)
- B \(2020\)
- C \(1576\)
- D \(1734\)
Answer & Solution
Correct Answer
(A) \(1818\)
Step-by-step Solution
Detailed explanation
\(\mathrm{PV} ^\gamma=\) constant \(\mathrm{TV} ^{\gamma-1}=\mathrm{C}\) \(300 \times \mathrm{V}^{\frac{7}{5}-1}=\mathrm{T}_{2}\left(\frac{\mathrm{V}}{16}\right)^{\frac{7}{5}-1}\) \(300 \times 2^{4 \times \frac{2}{5}}=\mathrm{T}_{2}\) Isobaric process…
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