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PART - 2 CH - 11 Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 11 Thermodynamics3 Marks
Question
By suddenly compressing a gas at a temperature of 300 K, its pressure is made 8 times the initial pressure. Calculate the temperature increase due to compression $(\gamma=1.5)$.
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Answer

$\begin{array}{l}T_1=300 K \\\text {Assume}\quad P_1=P \text { and } P_2=8 P\end{array}$
The increase in temperature due to compression is assumed to be $T_2$ is the adiabatic process.
$\begin{array}{ll}\therefore & P_1^{1-\gamma} T_1^\gamma=P_2^{1-\gamma} T_2^\gamma \text { constant } \\& \left(\frac{P_1}{P_2}\right)^{1-\gamma}=\left(\frac{T_2}{T_1}\right)^\gamma\end{array}$
Putting the value :
$\begin{array}{l}\left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}}=\frac{T_2}{T_1} \\\left(\frac{P}{8 P}\right)^{\frac{1-1.5}{1.5}}=\left(\frac{T_2}{300}\right) \\\left(\frac{1}{8}\right)^{\frac{-1}{3}}=\frac{T_2}{300} \\\left(\frac{1}{2}\right)^{-1}=\frac{T_2}{300} \Rightarrow 2=\frac{T_2}{300} \\\therefore T_2=600 K
\end{array}$
Increase in temperature due to compression
$\begin{array}{l}=T_2-T_1 \\=600 K-300 K \\=300 K\end{array}$

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