Q. $1\ mm^{3}$ of gas is compressed at 1 atm pressure and temperature $27℃\ to\ 627℃.$ What is the pressure under adiabatic condition? $(\gamma for\ the\ gas = 1.5)$

A) $27 \times 10^{5}\ Nm^{- 2}$
B) $12 \times 10^{5}\ Nm^{- 2}$
C) $15 \times 10^{5}\ Nm^{- 2}$
D) $23 \times 10^{3}\ Nm^{- 2}$
Explanation

(a) For adiabatic change, $\frac{T^{\gamma}}{P^{\gamma - 1}} = constant$
$\left( \frac{T_{1}}{T_{2}} \right)^{\gamma} = \left( \frac{p_{1}}{p_{2}} \right)^{\gamma - 1\ }$
${\ \ \ \ \left( \frac{27 + 273}{627 + 273} \right)}^{1.5} = \left( \frac{1}{p_{2}} \right)^{0.5}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{1}{3} \right)^{3/2} = \left( \frac{1}{P_{2}} \right)^{1/2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{2} = 27\ atm$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ 27 \times 1.07 \times 10^{5}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 27 \times 10^{5}\ Nm^{- 2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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