JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 15)
A bicycle tyre is filled with air having pressure of $$270 ~\mathrm{kPa}$$ at $$27^{\circ} \mathrm{C}$$. The approximate pressure of the air in the tyre when the temperature increases to $$36^{\circ} \mathrm{C}$$ is
262 kPa
360 kPa
270 kPa
278 kPa
Explanation
$\mathrm{P}_{\text {in }}=270 \mathrm{kPa}, \mathrm{T}_{\text {in }}=27^{\circ} \mathrm{C}$
$=300 \mathrm{~K}$
$$ \mathrm{T}_{\text {final }}=36^{\circ} \mathrm{C}=309 \mathrm{~K} $$
Hence we can consider process to be isochoric volume constant
$\therefore P \propto T$
$$ \frac{P_{\text {in }}}{P_{f}}=\frac{T_{\text {in }}}{T_{f}} \Rightarrow P_{f}=278 ~\mathrm{kPa} $$
$=300 \mathrm{~K}$
$$ \mathrm{T}_{\text {final }}=36^{\circ} \mathrm{C}=309 \mathrm{~K} $$
Hence we can consider process to be isochoric volume constant
$\therefore P \propto T$
$$ \frac{P_{\text {in }}}{P_{f}}=\frac{T_{\text {in }}}{T_{f}} \Rightarrow P_{f}=278 ~\mathrm{kPa} $$
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