The tyre of a bicycle has a volume of 2 times | vedclass

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Question

(C) The total pressure inside the tyre when the rider is on the bicycle is the sum of the atmospheric pressure and the pressure exerted by the weight

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) The total pressure inside the tyre when the rider is on the bicycle is the sum of the atmospheric pressure and the pressure exerted by the weight of the rider and bicycle.Pressure $P = P_{	ext{atm}} + \frac{Mg}{A} = 10^5 + \frac{120 	imes 10}{24 	imes 10^{-4}} = 10^5 + 5 	imes 10^5 = 6 	imes 10^5 ~Nm^{-2}$.Using the ideal gas law $PV = nRT$,the volume $V_1$ that the air currently inside the tyre would occupy at atmospheric pressure $P_{	ext{atm}}$ is given by $P_1 V_1 = P_{	ext{atm}} V_{	ext{final}}$.$V_1 = \frac{P 	imes V_{	ext{tyre}}}{P_{	ext{atm}}} = \frac{6 	imes 10^5 	imes 2 	imes 10^{-3}}{10^5} = 12 	imes 10^{-3} ~m^3$.The initial volume of air already present in the tyre at atmospheric pressure is $V_0 = 0.75 	imes 2 	imes 10^{-3} = 1.5 	imes 10^{-3} ~m^3$.The volume of air to be added at atmospheric pressure is $\Delta V = V_1 - V_0 = (12 - 1.5) 	imes 10^{-3} = 10.5 	imes 10^{-3} ~m^3$.Given the pump volume per stroke is $V_{	ext{pump}} = 500 ~cm^3 = 500 	imes 10^{-6} ~m^3 = 0.5 	imes 10^{-3} ~m^3$.The number of strokes $N = \frac{\Delta V}{V_{	ext{pump}}} = \frac{10.5 	imes 10^{-3}}{0.5 	imes 10^{-3}} = 21$.

Process	Condition
$(I)$ Adiabatic	$(A)$ $\Delta W = 0$
$(II)$ Isothermal	$(B)$ $\Delta Q = 0$
$(III)$ Isochoric	$(C)$ $\Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$
$(IV)$ Isobaric	$(D)$ $\Delta U = 0$

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