A certain amount of gas of volume V at 27^{o} | vedclass

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Question

$P _{1}=2 	imes 10^{7} Pa$

$P _{1} V _{1}= P _{2} V _{2}$

Since $V_{2}=2 V_{1}$ Hence $P_{2}=P_{1} / 2$ (isothermal expansion)

$P _{2}=1 	i

Vedclass · Accepted Answer

$3.536 	imes 10^{6}\,Pa$

Vedclass · Answer

$P _{1}=2 	imes 10^{7} Pa$

$P _{1} V _{1}= P _{2} V _{2}$

Since $V_{2}=2 V_{1}$ Hence $P_{2}=P_{1} / 2$ (isothermal expansion)

$P _{2}=1 	imes 10^{7} Pa$

$P _{2}\left( V _{2}ight)^{\gamma}= P _{3}\left(2 V _{2}ight)^{\gamma}$

$P _{3}=\frac{1 	imes 10^{7}}{2^{1.5}}=3.536 	imes 10^{6}$

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 $\gamma=1.5$ )

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