The cross-sectional area of a piston is 8 tim | vedclass

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(B) Initial state: $V_1 = 2.4 	imes 10^{-3} \, m^3$,$P_1 = 1.0 	imes 10^5 \, Pa$,$T_1 = 300 \, K$.The change in volume is $\Delta V = A 	imes x = 8

Vedclass · Accepted Answer

$800$

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(B) Initial state: $V_1 = 2.4 	imes 10^{-3} \, m^3$,$P_1 = 1.0 	imes 10^5 \, Pa$,$T_1 = 300 \, K$.The change in volume is $\Delta V = A 	imes x = 8 	imes 10^{-3} \, m^2 	imes 0.1 \, m = 0.8 	imes 10^{-3} \, m^3$.Final volume: $V_2 = V_1 + \Delta V = 2.4 	imes 10^{-3} + 0.8 	imes 10^{-3} = 3.2 	imes 10^{-3} \, m^3$.The pressure exerted by the spring on the piston is $P_{spring} = \frac{F}{A} = \frac{kx}{A} = \frac{8000 	imes 0.1}{8 	imes 10^{-3}} = \frac{800}{8 	imes 10^{-3}} = 1.0 	imes 10^5 \, Pa$.Final pressure: $P_2 = P_1 + P_{spring} = 1.0 	imes 10^5 + 1.0 	imes 10^5 = 2.0 	imes 10^5 \, Pa$.Using the ideal gas law for a fixed amount of gas: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.$\frac{1.0 	imes 10^5 	imes 2.4 	imes 10^{-3}}{300} = \frac{2.0 	imes 10^5 	imes 3.2 	imes 10^{-3}}{T_2}$.$T_2 = \frac{2.0 	imes 10^5 	imes 3.2 	imes 10^{-3} 	imes 300}{1.0 	imes 10^5 	imes 2.4 	imes 10^{-3}} = \frac{2.0 	imes 3.2 	imes 300}{2.4} = \frac{6.4 	imes 300}{2.4} = \frac{1920}{2.4} = 800 \, K$.

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