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The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)
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The correct answer is:
7
$\begin{aligned}
T_1=15^{\circ} \mathrm{C} & =15+273=288 \mathrm{~K} \\
T_2 & =35^{\circ} \mathrm{C}=35+273=308 \mathrm{~K}
\end{aligned}$
Volume remains constant.
So,
$\begin{aligned}
& \frac{p_1}{T_1}=\frac{p_2}{T_2} \\
& \frac{p_1}{P_2}=\frac{T_1}{T_2} \Rightarrow \frac{p_1}{P_2}=\frac{288}{308} \\
& \frac{p_2}{p_1}=\frac{308}{288}
\end{aligned}$
$\begin{aligned}
\% \text { increase in pressure } & =\frac{p_2-p_1}{p_1} \times 100 \\
& =\frac{308-288}{288} \times 100 \\
& \approx 7 \%
\end{aligned}$
T_1=15^{\circ} \mathrm{C} & =15+273=288 \mathrm{~K} \\
T_2 & =35^{\circ} \mathrm{C}=35+273=308 \mathrm{~K}
\end{aligned}$
Volume remains constant.
So,
$\begin{aligned}
& \frac{p_1}{T_1}=\frac{p_2}{T_2} \\
& \frac{p_1}{P_2}=\frac{T_1}{T_2} \Rightarrow \frac{p_1}{P_2}=\frac{288}{308} \\
& \frac{p_2}{p_1}=\frac{308}{288}
\end{aligned}$
$\begin{aligned}
\% \text { increase in pressure } & =\frac{p_2-p_1}{p_1} \times 100 \\
& =\frac{308-288}{288} \times 100 \\
& \approx 7 \%
\end{aligned}$
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