2.25 moles of an ideal gas with CV, m = 3R / 2 undergoes the transformations des
ID: 2258158 • Letter: 2
Question
2.25 moles of an ideal gas with CV, m = 3R / 2 undergoes the transformations described in the following list from an initial state described by T = 310 K and P = 1.00 bar. Calculate q, w, Delta U, Delta H, and Delta S for each process. The gas is heated to 675 K at a constant external pressure of 1.00 bar. The gas is heated to 675 K at a constant volume corresponding to the initial volume. The gas undergoes a reversible isothermal expansion at 310 K until the pressure is one third of its initial value.Explanation / Answer
(a) delta U = 2.25*3R/2 * (675-310) = 10.241 KJ
W = P*deltaV = nr*delta T = 6.87KJ
Q = U+W = 17.68KJ
(b) delta U = 10.241KJ
W = zero
Q = U+W = 10.241KJ
(c) U = zero as this is isothermal
work = Pv/3 = nr delta T /3 = 2.29
Q = 2.29
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