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During an adiabatic reversible expansion of an ideal diatomic gas, initially the pressure was $2$ atm and volume was $5$ L, after the expansion the volume occupied by gas was $15$ L. What is the pressure in the container after expansion?
Option: 1
$2\times\left ( \frac{1}{3}\right )^{\frac{7}{5}}\mathrm{atm}$

Option: 2
$2\times\left ( \frac{1}{3}\right )^{\frac{5}{3}}\mathrm{atm}$

Option: 3
$(\frac{2}{3})\: \mathrm{atm}$

Option: 4
$\left ( \frac{2}{3} \right )^{\frac{5}{3}}\mathrm{atm}$

Answers (1)

best_answer

During an adiabatic reversible process, the $PV$ relationship is given by:

$PV^{\gamma }=$ constant

For a diatomic gas value of = $7/5$

$P_{1}V_{1}^{\gamma }=P_{2}V_{2}^{\gamma }$

$2\times 5^{\left ( \frac{7}{5} \right )}=P_{2}\times 15^{\left ( \frac{7}{5} \right )}$

$P_{2}=2\times \left ( \frac{1}{3} \right )^{\left ( \frac{7}{5} \right )}\mathrm{atm}$
Hence, option number (1) is correct

Posted by

HARSH KANKARIA

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