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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
The pressure of the wheel of motor car at any instant is equal to 2 atmospheric pressure at the temperature is $15^{\circ} C$. At this moment the tube of the wheel bursts. Find the decrease in temperature of the released $\operatorname{air}(\gamma=1.4)$
✓

Answer

Given that :
$P_1=2 Atmospheric$
$\begin{aligned}T_1 & =15^{\circ} C=15+273=288 K \\\gamma & =1.4 \\T_2 & =? \\P_2 & =1 \text { Atmospheric }\end{aligned}$
And
We know that
\begin{array}{l} 
P_1^{1-\gamma} T_1^\gamma=P_2^{1-\gamma} T_2^\gamma=\text { Constant } \\\begin{aligned}\left(\frac{P_1}{P_2}\right)^{1-\gamma} & =\left(\frac{T_2}{T_1}\right)^\gamma \\\left(\frac{T_2}{T_1}\right)^\gamma & =\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \\T_2 & =T_1\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \\& =288\left(\frac{1}{2}\right)^{\frac{1.4-1}{1.4}} \\& =288\left(\frac{1}{2}\right)^{2 / 7} \\
T_2 & =\frac{280}{2^{2 / 7}}\end{aligned}\end{array}
Taking log both sides :
$\begin{aligned}\log T_2 & =\log 288-\frac{2}{7}\log 2 \\& =2.4594-\frac{2}{7}(0.3010) \\& =2.4594-0.086\\\log T_2 & =2.3734 \\ T_2 & =\text { Anti } \log 2.3734\\\therefore\quad T_2 & =236.3 K\end{aligned}$
Decrease in temperature
$\begin{array}{l}=288-236.3 \\=51.7 K\end{array}$

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