(D) Given: Initial pressure $P_1 = 270 \text{ kPa}$,Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.Final temperature $T_2 = 37^{\c

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(D) Given: Initial pressure $P_1 = 270 \text{ kPa}$,Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.Final temperature $T_2 = 37^{\circ}C = 37 + 273 = 310 \text{ K}$.Assuming the volume of the tyre remains constant,according to Gay-Lussac's Law,$\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Substituting the values: $\frac{270}{300} = \frac{P_2}{310}$.$P_2 = \frac{270 \times 310}{300} = 0.9 \times 310 = 279 \text{ kPa}$.Therefore,the final pressure is $279 \text{ kPa}$.

Class 11 Physics Kinetic Theory of Gases Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

Easy

$A$ tyre of a vehicle is filled with air having pressure $270 \text{ kPa}$ at $27^{\circ}C$. The air pressure in the tyre when the temperature increases to $37^{\circ}C$ is (in $\text{ kPa}$)

MHT CET 2025 All MHT CET

A
$282$
B
$270$
C
$265$
D
$279$

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Class 11

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Kinetic Theory of Gases

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Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

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$A$ tyre of a vehicle is filled with air having pressure $270 \text{ kPa}$ at $27^{\circ}C$. The air pressure in the tyre when the temperature increases to $37^{\circ}C$ is (in $\text{ kPa}$)

Similar Questions

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