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Explanation & Answer
the heat cacpacity of 4.1 mole Argon is
C = 3/2 n R = 3/2*4.1*8.31 = 51.10J/K
b) 960/51.10 = 18.78 K
c)
PV = nRT
P deltaV = nR delta T
deltaV = nR/P *deltaT = 4.1*8.31/120000*18.78
=0.00533 m^3
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