The tyre of a motor car contains air at 15^{c | vedclass

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(A) Given:Initial temperature $T_1 = 15^{\circ} C = 15 + 273 = 288 \ K$.Final temperature $T_2 = 35^{\circ} C = 35 + 273 = 308 \ K$.Since the volume o

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$7$

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(A) Given:Initial temperature $T_1 = 15^{\circ} C = 15 + 273 = 288 \ K$.Final temperature $T_2 = 35^{\circ} C = 35 + 273 = 308 \ K$.Since the volume of the tyre remains constant, we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Rearranging for the ratio of pressures: $\frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308}{288}$.The percentage increase in pressure is given by $\frac{P_2 - P_1}{P_1} 	imes 100 = \left( \frac{P_2}{P_1} - 1 ight) 	imes 100$.Substituting the values: $\left( \frac{308}{288} - 1 ight) 	imes 100 = \left( \frac{308 - 288}{288} ight) 	imes 100 = \frac{20}{288} 	imes 100 \approx 6.94 \%$.Rounding to the nearest integer, the approximate percentage increase is $7 \%$.

The tyre of a motor car contains air at $15^{\circ} C$. If the temperature increases to $35^{\circ} C$, the approximate percentage increase in pressure is (ignore the expansion of the tyre).

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