LON-CAPA p d v doubled x C N-235 Mol of Hydroge... x How To Take Screenshots X a
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LON-CAPA p d v doubled x C N-235 Mol of Hydroge... x How To Take Screenshots X a Search or enter address v e a Search Course Contents Homework set 10 (due 12/8 at 11:59PM p and V doubled 2PDH------------ 2 V O2003 Thomson Brooks/Cole A quantity of a monatomic and volume are doubled as shown in deal gas undergoes a process in which both its pressure the figure above. DATA: Vo- 0.35 m3 Po 9500 Pl What is the change of the internal energy of the gas? Submit Answer Tries 0/20 What was the work done by the gas during the expansion? Submit Answer Tries 0/20 What amount of heat flowed into the gas during the expansion? Sub Answer Tries 0/20 38 Post Discussion HE Search the web and Windows Timer Notes Evaluate Feedback Print Info Send Feedback 9.45 AM 12/8/2015Explanation / Answer
U = Q - W
Q = U + W
use ideal gas law:
p0V0 = nRT0 and
p1V1 = nRT1
---> T1 = p1V1/(nR)
with p1 = 2p0 and V2 = 2V1
T1 = 2p0*2V0/(nR) = 4p0V0/(nR)
T = T1 - T0 = 4p0V0/(nR) - p0V0/(nR) = 3p0V0/nR
And U = Cv*T with Cv for an ideal gas = 3/2*R
So U = n*3/2*R*3p0V0/(nR) = 4.5p0V0
U = 4.5*9500*0.35 J = 14962.5 J (change of internal energy)
Work done is
W = 1/2(p0+p1)(V1-V0) = 1/2(p0+2p0)(2V0-V0)
= 1/2(3p0)(V0) = 1.5p0V0
W = 1.5*9500*0.35 = 4987.5J (work done)
Q = U + W =14962.5 + 4987.5 19950 J (heat flow into gas)
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