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A gas at $37^{\circ} \mathrm{C}$ is compressed adiabatically to half of its volume, then the final temperature of the gas is
(Ratio of specific heat capacities of the gas is 1.5)
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(Ratio of specific heat capacities of the gas is 1.5)
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Verified Answer
The correct answer is:
$165.3^{\circ} \mathrm{C}$
In an Adiabatic process,
$T V^{\gamma-1}=$ constant
$$
\Rightarrow T_1 \cdot V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}
$$
Here, $r=1.5, V_2=V_1 / 2$
and $T_1=37^{\circ} \mathrm{C}=310.15 \mathrm{~K}$
So, final temperature is $T_2=15 \times(2)$
$$
\begin{aligned}
& =310.15 \times \sqrt{2} \\
& =437.31 \mathrm{~K}=164.12^{\circ} \mathrm{C} \\
& \simeq 165.3^{\circ} \mathrm{C}
\end{aligned}
$$
$T V^{\gamma-1}=$ constant
$$
\Rightarrow T_1 \cdot V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}
$$
Here, $r=1.5, V_2=V_1 / 2$
and $T_1=37^{\circ} \mathrm{C}=310.15 \mathrm{~K}$
So, final temperature is $T_2=15 \times(2)$
$$
\begin{aligned}
& =310.15 \times \sqrt{2} \\
& =437.31 \mathrm{~K}=164.12^{\circ} \mathrm{C} \\
& \simeq 165.3^{\circ} \mathrm{C}
\end{aligned}
$$
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