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

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 11 Thermodynamics2 Marks
Question
A gas having volume of 1 liter at 80 cm mercury pressure expands to 1.190 liter in the adiabatic process. If the pressure of the gas is $6 2 . 6 ~ c m$ and it falls to then find the value of $\gamma$.
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Answer

Given that :
$\begin{aligned}P_1 & =80 cm \\P_2 & =62.6 cm \\V_1 & =1 \text { litre } \\V_2 & =1.190 \text { litre }\end{aligned}$
For adiabatic change
$P _1 V_1^\gamma= P _2 V_2^\gamma$
or $\quad\left(\frac{ V _1}{V_2}\right)^\gamma=\left(\frac{ P _2}{ P _1}\right)$
$\begin{aligned}\text {or}\quad\quad\gamma\left(\log V_{1}-\log V_2\right) & =\log P_2-\log P_1 \\\gamma & =\frac{\log P_2-\log P_1}{\log V_1-\log V_2} \\& =\frac{\log 62.6-\log 80}{\log 1-\log 1.190} \\& =\frac{1.7966-1.9031}{0-0.0755}=\frac{0.1065}{0.0755} \\\gamma & =1.41\end{aligned}$

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