A cylindrical tube of cross-sectional area A | vedclass

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(C) Since the pistons are connected by a wire,the volume $V$ of the gas remains constant.Using the ideal gas law for a constant volume process: $\frac

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$P_0 A$

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(C) Since the pistons are connected by a wire,the volume $V$ of the gas remains constant.Using the ideal gas law for a constant volume process: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Given $P_1 = P_0$,$T_1 = T_0$,and $T_2 = 2T_0$,we have:$\frac{P_0}{T_0} = \frac{P'}{2T_0} \Rightarrow P' = 2P_0$.The pressure inside the tube is now $2P_0$,while the external atmospheric pressure is $P_0$.The net force acting outwards on each piston is $F_{net} = (P' - P_0)A = (2P_0 - P_0)A = P_0 A$.Since the wire connects the two pistons,the tension $T$ in the wire must balance this net outward force.Therefore,the tension in the wire is $T = P_0 A$.

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