Work done=pex.(∆V)=1.2 kJ=1200 J=1.02atm( Vf-0.28)
1200=1.02×101.33J(Vf-0.28)
Vf=11.89 L
Tf=Vf.Pf/ R=11.89×1.02/0.082=147.9 K
Ti=2+273=275 K.
∆T=147.9-275=-127.1 K
Cp=Cv+R=5/2 R
∆H=nCp.∆T
=5/2R×-127.1
=-2641.8 J
Incorrect Question 8 0/1 pts 0.28 litre of an ideal monatomic gas (Cv.m 3R/2) initially at...
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