An ideal gas at a pressure $1\ atm$ and temperature of $27℃$ is compressed adiabatically until its pressure becomes 8 times the initial pressures. Then the final temperature is $\left( \gamma = \frac{3}{2} \right)$

Question

Dr. Abubacker · Accepted Answer

(d) Here, $p = 1$ atm, $T_{1} = 27℃$
$= 27 + 273 = 300$ K
$p_{2} = 8$ atm, $T_{2} = ?,\ \gamma = 3/2$
As changes are adiabatic,
$	herefore\ \ p_{1^{\gamma - 1}T_{1}^{- \gamma} = P_{2}^{\gamma - 1}T_{2}^{- \gamma}}$
$\left( \frac{T_{2}}{T_{1}} ight)^{- \gamma} = \left( \frac{p_{1}}{p_{2}} ight)^{\gamma - 1}$
$T_{2} - T_{1}\left( \frac{p_{1}}{p_{2}} ight)^{\gamma - 1/\gamma} = 300(8)^{(1.5 - 1)/1.5} = 300(8)^{1/3}$
$T = 600K = (600 - 273)℃ = 327℃$

Q. An ideal gas at a pressure $1\ atm$ and temperature of $27℃$ is compressed adiabatically until its pressure becomes 8 times the initial pressures. Then the final temperature is $\left( \gamma = \frac{3}{2} \right)$

Explanation

Reviewed by Dr. Abubacker

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