An ideal monoatomic gas is confined in a cyli | vedclass

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(C) Given:Cross-sectional area $A = 8.0 	imes 10^{-3} \, m^2$Initial temperature $T_1 = 300 \, K$Initial volume $V_1 = 2.4 	imes 10^{-3} \, m^3$Disp

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(C) Given:Cross-sectional area $A = 8.0 	imes 10^{-3} \, m^2$Initial temperature $T_1 = 300 \, K$Initial volume $V_1 = 2.4 	imes 10^{-3} \, m^3$Displacement of piston $\Delta x = 0.1 \, m$Spring constant $k = 8000 \, N/m$Atmospheric pressure $P_0 = 1.0 	imes 10^5 \, N/m^2$$1$. Calculate the final volume $V_2$:$V_2 = V_1 + A \Delta x = 2.4 	imes 10^{-3} + (8.0 	imes 10^{-3} 	imes 0.1) = 2.4 	imes 10^{-3} + 0.8 	imes 10^{-3} = 3.2 	imes 10^{-3} \, m^3$$2$. Calculate the final pressure $P_2$:The final pressure is the sum of atmospheric pressure and the pressure exerted by the spring force:$P_2 = P_0 + \frac{k \Delta x}{A} = 1.0 	imes 10^5 + \frac{8000 	imes 0.1}{8.0 	imes 10^{-3}} = 1.0 	imes 10^5 + \frac{800}{8.0 	imes 10^{-3}} = 1.0 	imes 10^5 + 1.0 	imes 10^5 = 2.0 	imes 10^5 \, N/m^2$$3$. Use the ideal gas law $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$:Initial pressure $P_1 = P_0 = 1.0 	imes 10^5 \, N/m^2$ (since the spring is initially relaxed).$\frac{1.0 	imes 10^5 	imes 2.4 	imes 10^{-3}}{300} = \frac{2.0 	imes 10^5 	imes 3.2 	imes 10^{-3}}{T_2}$$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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