Starting at temperature 300 ; K, one mole of an ideal diatomic ga… - Vidyadip

MCQ

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)

✓
$1818$
B
$2020$
C
$1576$
D
$1734$

✓

Answer

Correct option: A.

$1818$

a
$\mathrm{PV} ^\gamma=$ 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

$\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}$

$\mathrm{V}_{2}=\mathrm{kT}_{2}$

$2 \mathrm{V}_{2}=\mathrm{KT}_{\mathrm{f}}$

$\frac{1}{2}=\frac{\mathrm{T}_{2}}{\mathrm{T}_{\mathrm{f}}} \Rightarrow \mathrm{T}_{\mathrm{f}}=2 \mathrm{T}_{2}$

$\mathrm{T}_{\mathrm{f}}=2 \times 300 \times 2^{\frac{8}{5}}=1818.85$

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