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Two moles of an ideal monoatomic gas undergo expansion following the relation P*T = constant, starting from initial state (P0, V0) until the pressure becomes P0/2. Find the change in internal energy of the gas.

1. 3/4 * P0*V0
2. 3/2 * P0*V0
3. 9/2 * P0*V0
4. 5/2 * P0*V0

Correct answer: 3/2 * P0*V0

Solution

PT = constant = P0*T0. From PV = nRT: T = PV/(nR). So P*(PV/nR) = const => P²*V = const. Initial: P0²*V0. Final: (P0/2)² * Vf = P0²*V0 => Vf = 4V0. T0 = P0*V0/(nR), Tf = (P0/2)*(4V0)/(nR) = 2P0*V0/(nR) = 2T0. delta_U = n*(3/2)*R*(Tf - T0) = 2*(3/2)*R*(T0) = 3R*T0 = 3*(P0*V0/n)*n... R*T0 = P0*V0/n (since n=2: R*T0 = P0*V0/2). So delta_U = 3*R*T0 = 3*(P0*V0/2) = 3P0*V0/2.

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