n the figure, 2.62 mole of an ideal diatomic gas can go from a to c along either
ID: 1404008 • Letter: N
Question
n the figure, 2.62 mole of an ideal diatomic gas can go from a to c along either the direct (diagonal) path ac or the indirect path abc. The scale of the vertical axis is set by pab 4.89 Pa and p 2.20 kPa, and the scale of the horizontal axis is set by Vb 5.35 m3 and V. 2.49 m The molecules rotate but do not oscillate.) During the transition, (a) what is the change in internal energy of the gas, and (b) how much energy is added to the gas as heat? (c) How much heat is required if the gas goes from a to c along the indirect path abc? Pab Volume (mExplanation / Answer
a)
at point a ,
PV = n * R * Ta
4.89 *10^3 * 2.49 = 2.62 * 8.314 * Ta
Ta = 559 K
Now, at point C
PV = n * R * Tc
2.2 *10^3 * 5.35 = 2.62 * 8.314 * Tc
Tc = 540.33 K
NOw, change in internal energy = 2.5 * 2.62 ** 8.314 (540.33 - 559)
change in internal energy = -1016.3 J
the change in internal energy is -1016.3 J
b)
work done by the gas = area under the curve
work done by the gas = (4.89 - 2.20) * 10^3 * 0.5 * (5.35 - 2.49) + (5.35 - 2.49) * 2.20 * 10^3
work done by the gas = 10138.7 J
heat added = work done by the gas + change in internal energy
heat added = 10138.7 - 1016.3
heat added = 9122.4 J
the heat added is 9122.4 J
c)
for the indirect path ,
work dobe by the gas = (5.35 - 2.49) 4.89 * 10^3
work dobe by the gas = 13895.4 J
heat added = work done by the gas + change in internal energy
heat added = 13895.4 J - 1016.3
heat added = 12969 J
the heat added is 12969 J
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