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Two moles of Helium gas (γ = 5/3) are initially at temperature 27oC and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. What are the final volume?
(A) 113.13 × 10–3 m3
(B) 213.13 × 10–3 m3
(C) 313.13 × 10–3 m3
(D) 13.13 × 10–3 m3
pv = nRT
for first process pressure is kept constant so
v directionally proportional to temp
Vi/Vf = Ti/Tf
Vf = 2Vi , Vi = 20L & Ti=300k
so T2 = 600k
now second process is adiabatic so TVr-1 = c
Ti/Tf = (Vf/Vi)r-1
Ti = 600 ,Tf=300 , Vi=20*2=40L & r=5/3 so
Vf = 80*21/2 L
=113.13*10-3 m3
thank you
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