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PHYS1131finalS12010SOL2

# PHYS1131finalS12010SOL2 - HigherPhysics1A Q3...

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Higher Physics 1A PHYS1131 2010-T1, Solutions Q3 27 Marks (a) Work done on the gas W = PdV V 1 V 2 where pressure and volume are P and V respectively. V 1 is initial volume and V 2 is final volume. NOTE THE MINUS SIGN (b) (i) When a closed system undergoes a changes in the state the change in internal energy of the system is given by: Δ E int = Q + W where Q is the heat supplied to the system W is the work done on the system and Δ E int is a state function (ii) For a cyclic system Δ E int = 0 as it returns to the starting point, so that Q+W=0. (c) (i) Work done W = P Δ V = P ( V f V i ) where P is a constant and V i , V f the initial and final volumes. For an ideal gas PV = nRT where n is the # moles. V f = nRT f P and T f is the final temperature. For the liquid water V i = Mass Density = nM H 2 O ρ water where M H2O is the molar mass of water. W = P nRT f P nM H 2 O ρ = nRT f + nPM H 2 O ρ = ‐[2.00 . 8.314 . (273.16+100)] + [2.00 . 1.013x10 5 . 18/1000]/(1.0 x 10 6 ) J =‐6204.9 + 3.6 10 ‐3 J =‐6204.9 J (i.e. the volume of the liquid water is negligible and can be neglected) =‐6.20 x 10 3 J to 3SF (ii) Change in internal energy is given by the First Law; i.e. Δ E int = Q + W with Q being heat supplied, W being the work done. Q = m water L W with L W the latent heat of fusion = 2.00 18 10 3 2.30 10 6 J = 82,800 J Thus, Δ E = 82,800 6,201 J = 76,599 J = 7.66 × 10 4 J to 3SF

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(d) (i) We know from the First Law of Thermodynamics that Δ E int = Q + W . We have Δ E AC = 800J, W ABC = 500 J = W AB + W BC From conservation of energy Δ E ABC = Δ E AC since the same initial and final states.
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