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A certain amount of gas of volume $\mathrm{V}$ at $27^{\circ} \mathrm{C}$ temperature and pressure $2 \times 10^{7} \mathrm{Nm}^{-2}$ 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 $\gamma=1.5$ ) :

Option: 1
$3.536 \times 10^{5} \mathrm{~Pa}$

Option: 2
$3.536 \times 10^{6} \mathrm{~Pa}$

Option: 3
$1.25 \times 10^{6} \mathrm{~Pa}$

Option: 4
$1.25 \times 10^{5} \mathrm{~Pa}$

Answers (1)

best_answer

Process $\mathrm{A\rightarrow B}$ is isothermal process

$\therefore \mathrm{P_{1}V=P_{2}\left ( 2v \right )}$

$\mathrm{P_{2}=\frac{P_{1}}{2}}\rightarrow (1)$

Process $\mathrm{B\rightarrow C}$ is adiabatic process

$\mathrm{P_2(2 v)^v=P_3(4 v)^v }$

$\mathrm{\frac{P_2}{P_3}=(2 )^{v}=(2 )^3/2 }$

$\mathrm{\frac{P_2}{P_3}=2 \sqrt{2}\: \: \rightarrow 2}$

From eq.(1)&(2)

$\mathrm{\frac{P_1}{2} =2 \sqrt{2} \: \: P_3 }$

$\mathrm{P_3 =\frac{P_1}{4 \sqrt{2}} }$

$\mathrm{=\frac{2 \times 10^7}{4 \sqrt{2}} \\ }$

$\mathrm{P_3 =0.3536 \times 10^7}$

The final pressure of the gas will be $=3.536 \times 10^6\: \: \mathrm{pa}$

Hence (2) is correct option.

Posted by

Gautam harsolia

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