1.

Two moles of an ideal monoatomic gas occupies a volume V at 27^@ C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change on its internal energy.

Answer»

(a)189 K (b)2.7 kJ
(a)195 K (b)-2.7 kJ
(a)189 K (b)-2.7 kJ
(a)195 K (b)2.7 kJ

Solution :For adiabatic CHANGE ,
`PV^gamma` =constant and equation of state of gas
`PV=muRT`
`therefore (PV^gamma)/(PV)="constant"/(muRT)`
`therefore V^(gamma-1) = "constant"/T` [`because mu R`= constant]
`therefore V^(gamma-1) PROP 1/T`
`therefore (V_1/V_2)^(gamma-1)=T_2/T_1`
`therefore T_2=T_1xx(V_1/V_2)^(gamma-1)`
`=300xx(1/2)^(5/3-1)`
`=300xx(0.5)^(2/3)`
=189 K
Change in internal energy ,
`DeltaU=(mu f R DeltaT)/2`
`=2xx3/2xx8.314xx(189-300)`
=3 x 8.314 x (-111)
=- 333 x 8.314
=-2768 J
`therefore DeltaU APPROX` -2.7 KJ (almost near value)


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