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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 is suddenly compressed from its original volume to a volume of $1 / 4$. Estimate the temperature, the original temperature is 300 K and $\gamma= 1 . 5$.
✓

Answer

 $ \begin{aligned}\text {Suppose}\quad\quad V_1 & =\text { Initial volume } \\ V_2 & =\text { Final volume }=\frac{V_1}{4} \\ \frac{V_1}{V_2} & =4 \quad \therefore \frac{V_2}{V_1}=\frac{1}{4} \\ T_1 & =300 K, T_2=? \\ \gamma & =1.5 \end{aligned} $
We know that for adiabatic change :
$\begin{aligned}T _1 V_1{ }^{\gamma-1} & = T _2 V_2^{\gamma-1} \\ \text{or}\quad \frac{T_1}{T_2} & =\left(\frac{ V _2}{V_1}\right)^{\gamma-1} \\\text {or}\quad \frac{300}{T_2} & =\left(\frac{1}{4}\right)^{1.5-1}=\left(\frac{1}{4}\right)^{0.5} \\ \text {or}\quad\frac{300}{T_2} & =\frac{1}{2} \quad \therefore T_2=600 K\end{aligned}$
Hence the increase in temperature
$\begin{array}{l}=T_2-T_1 \\ =600-300=300 K\end{array}$

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